G
E
PR
O
O
FS
11
LI
N
E
PA
Exponential
functions
11.1 Kick off with CAS
O
N
11.2 Indices as exponents
11.3 Indices as logarithms
11.4 Graphs of exponential functions
11.5 Applications of exponential functions
11.6 Inverses of exponential functions
11.7 Review
11.1 Kick off with CAS
Exponential functions
1 Using CAS technology, sketch the following exponential functions on the same
set of axes.
a y = 2x
5
6
9
O
N
LI
N
E
10
PA
8
G
E
7
O
FS
4
x
O
3
x
PR
2
x
1
b y=3
c y=5
d y=8
e y=
2
x
Using CAS technology, enter y = a into the function entry line and use a slider to
change the value of a.
When sketching an exponential function, what is the effect of changing the value
of a in the equation?
Using CAS technology, sketch the following exponential functions on the same
set of axes.
a y = 2x
b y = 2x + 2
c y = 2x + 5
d y = 2x − 3
e y = 2x − 5
Using CAS technology, enter y = 2x + k into the function entry line and use a
slider to change the value of k.
When sketching an exponential function, what is the effect of changing the value
of k in the equation?
Using CAS technology, sketch the following exponential functions on the same
set of axes.
a y = 2x
b y = 2x−2
c y = 2x+3
d y = 2x−5
e y = 2x−8
Using CAS technology, enter y = 2x−h into the function entry line and use a slider
to change the value of h.
When sketching an exponential function, what is the effect of changing the value
of h in the equation?
On the one set of axes, sketch the graphs of y1 = 2x and y2 = −2x−2 + 2.
Describe the transformations to get from y1 to y2.
x
Please refer to the Resources tab in the Prelims section of your eBookPlUs for a comprehensive
step-by-step guide on how to use your CAS technology.
Units 1 & 2
AOS 1
Topic 8
Concept 1
Indices as
exponents
Concept summary
Practice questions
Index or exponential form
When the number 8 is expressed as a power of 2, it is written as 8 = 23. In this form, the
base is 2 and the power (also known as index or exponent) is 3. The form of 8 expressed
as 23 is known both as its index form with index 3 and base 2, and its exponential form
with exponent 3 and base 2. The words ‘index’ and ‘exponent’ are interchangeable.
For any positive number n where n = ax, the statement n = ax is called an index or
exponential statement. For our study:
• the base is a where a ∈ R + \ 1
• the exponent, or index, is x where x ∈ R
• the number n is positive, so ax ∈ R + .
Index laws control the simplification of expressions which have the same base.
O
FS
11.2
Indices as exponents
Review of index laws
am × an = am+n
PR
am ÷ an = am−n
O
Recall the basic index laws:
(am) n = amn
From these, it follows that:
E
a0 = 1
1
1
and −n = an
n
a
a
G
a−n =
PA
(ab) n = anbn
−n
a
b
E
LI
N
n
a
b
=
an
bn
=
b
a
n
Fractional indices
O
N
Since a
1 2
2
1
= a and ( a) 2 = a, then a2 =
a. Thus, surds such as
in index form as 3 .
n
The symbols
, 3 , ...
are radical signs. Any radical can be converted to and
from a fractional index using the index law:
1
an =
n
a
m
n
m
n
A combination of index laws allows a to be expressed as a = a
m
n
can also be expressed as a = am
m
an =
572 3 can be written
1
2
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
1
n
=
n
n
a
am .
m
m
or a n =
n
am
1 m
n
n
= ( a) m. It
WORKeD
exaMpLe
1
a Express
32n × 91−n
as a power of 3.
81n−1
5 2
b Simplify
3a2b2
× 2(a−1b2) −2.
2
3
−
4
c Evaluate 8 +
9
1
2
.
tHinK
WritE
a 1 Express each term with the same base.
2n
1−n
32n × (32) 1−n
a 3 ×9
=
(34) n−1
32n × 32(1−n)
=
34(n−1)
O
FS
81n−1
2 Apply an appropriate index law.
Note: To raise a power to a power multiply the indices.
32n × 32−2n
34n−4
32n+2−2n
= 4n−4
3
O
=
3 Apply an appropriate index law.
PR
Note: To multiply numbers with the same base, add
the indices.
=
E
PA
G
∴
b 1 Use an index law to remove the brackets.
b
34n−4
= 32− (4n−4)
= 36−4n
4 Apply an appropriate index law and state the answer.
Note: To divide numbers with the same base, subtract
the indices.
32
32n × 91−n
= 36−4n
81n−1
5 2
2 2
3a b
× 2(a−1b2) −2
= 32a4b5 × 2 × a2b−4
E
= 9a4b5 × 2a2b−4
2 Apply the index laws to terms with the same base to
simplify the expression.
= 18a6b1
LI
N
= 18a6b
2
c 83 +
c 1 Express each term with a positive index.
N
Note: There is no index law for the addition
of numbers.
4
9
1
2
−
2
O
=
2 Apply the index law for fractional indices.
9
4
= 83 +
3
8
2
+
Note: The fractional indices could be interpreted in
2
1
2
9
4
2
other ways including, for example, as 83 = (23) 3.
3 Evaluate each term separately and calculate the
answer required.
= (2) 2 +
=4+1
=5
1
2
1
2
3
2
topic 11 expOnentIaL FunCtIOns
573
Indicial equations
An indicial equation has the unknown variable as an exponent. In this section we
shall consider indicial equations which have rational solutions.
Method of equating indices
If index laws can be used to express both sides of an equation as single powers of the
same base, then this allows indices to be equated. For example, if an equation can
be simplified to the form 23x = 24, then for the equality to hold, 3x = 4. Solving this
linear equation gives the solution to the indicial equation as x = 43.
O
FS
In the case of an inequation, a similar method is used. If 23x < 24, for example,
then 3x < 4. Solving this linear inequation gives the solution to the indicial
inequation as x < 43.
PR
O
To solve for the exponent x in equations of the form ax = n:
• Express both sides as powers of the same base.
• Equate the indices and solve the equation formed to obtain
the solution to the indicial equation.
E
Inequations are solved in a similar manner. However, you need to ensure the base
a is greater than 1 prior to expressing the indices with the corresponding order sign
between them.
1 x
2
3
< 12 as 2−x < 2−3 and then solve the linear
inequation −x < −3 to obtain the solution x > 3.
WORKeD
exaMpLe
2
PA
G
For example, you first need to write
Solve 53x × 254−2x =
E
tHinK
1
for x.
125
LI
N
1 Use the index laws to express the left-hand side of
N
the equation as a power of a single base.
O
2 Express the right-hand side as a power of the
same base.
WritE
1
53x × 254−2x = 125
1
53x × 52(4−2x) = 125
1
53x+8−4x = 125
1
58−x = 125
58−x = 513
58−x = 5−3
3 Equate indices and calculate the required value of x.
Equating indices,
8 − x = −3
x = 11
Indicial equations which reduce to quadratic form
The technique of substitution to form a quadratic equation may be applicable to
indicial equations.
574
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
To solve equations of the form p × a2x + q × ax + r = 0:
• Note that a2x = (ax) 2.
• Reduce the indicial equation to quadratic form by using a substitution for ax.
• Solve the quadratic and then substitute back for ax.
• Since ax must always be positive, solutions for x can only be obtained for ax > 0;
reject any negative or zero values for ax.
WORKeD
exaMpLe
3
Solve 32x − 6 × 3x − 27 = 0 for x.
WritE
O
FS
tHinK
32x − 6 × 3x − 27 = 0
Let a = 3x
∴ a2 − 6a − 27 = 0
1 Use a substitution technique to reduce the indicial
O
equation to quadratic form.
Note: The subtraction signs prevent the use of
index laws to express the left-hand side as a power
of a single base.
(a − 9)(a + 3) = 0
a = 9, a = −3
PR
2 Solve the quadratic equation.
G
E
3 Substitute back and solve for x.
Replace a by 3x.
∴ 3x = 9 or 3x = −3 (reject negative value)
3x = 9
∴ 3 x = 32
∴x=2
PA
scientific notation (standard form)
O
N
LI
N
E
Index notation provides a convenient way to express numbers which are either very
large or very small. Writing a number as a × 10b (the product of a number a where
1 ≤ a < 10 and a power of 10) is known as writing the number in scientific notation
(or standard form). The age of the earth since the Big Bang is estimated to be
4.54 × 109 years, while the mass of a carbon atom is approximately 1.994 × 10−23
grams. These numbers are written in scientific notation.
To convert scientific notation back to a basic numeral:
• move the decimal point b places to the right if the power of 10 has a
positive index, in order to obtain the large number a × 10b represents;
or
• move the decimal point b places to the left if the power of 10 has a
negative index, in order to obtain the small number a × 10−b represents.
This is because multiplying by 10−b is equivalent to dividing by 10b.
significant figures
When a number is expressed in scientific notation as either a × 10b or a × 10−b,
the number of digits in a determines the number of significant figures in the
basic numeral. The age of the Earth is 4.54 × 109 years in scientific notation or
4 540 000 000 years to three significant figures. To one significant figure, the age
would be 5 000 000 000 years.
topic 11 expOnentIaL FunCtIOns
575
WORKeD
exaMpLe
4
a Express each of the following numerals in scientific notation and state the
number of significant figures each numeral contains.
i 3 266 400
ii 0.009 876 03
b Express the following as basic numerals.
ii 1.037 × 10−5
i 4.54 × 109
tHinK
WritE
a i 1 Write the given number as a value between
a i In scientific notation,
3 266 400 = 3.2664 × 106
O
FS
1 and 10 multiplied by a power of 10.
Note: The number is large so the power of
10 should be positive.
2 Count the number of digits in the number a
There are 5 significant figures in the
number 3 266 400.
in the scientific notation form a × 10 and
state the number of significant figures.
1 and 10 multiplied by a power of 10.
Note: The number is small so the power of
10 should be negative.
ii 0.009 876 03 = 9.876 03 × 10−3
PR
ii 1 Write the given number as a value between
O
b
G
in the scientific notation form and state the
number of significant figures.
0.009 876 03 has 6 significant figures.
E
2 Count the number of digits in the number a
b i 4.54 × 109
b i 1 Perform the multiplication.
ii 2 Perform the multiplication.
PA
Note: The power of 10 is positive, so a
large number should be obtained.
ii 1.037 × 10−5
Move the decimal point 5 places to the left.
∴ 1.037 × 10−5 = 0.00001037
E
Note: The power of 10 is negative, so a
small number should be obtained.
LI
N
Move the decimal point 9 places to the
right.
∴ 4.54 × 109 = 4 540 000 000
ExErCisE 11.2 Indices as exponents
N
PraCtisE
1
O
Work without Cas
WE1
a Express
21−n × 81+2n
as a power of 2.
161−n
−2
1
2 −2
1
3 −4 2
b Simplify (9a b ) × 2 a b
c Evaluate 27
2 Simplify
−
2
3
20p5
m3q−2
+
÷
1
49 2
.
81
5(p2q−3) 2
−4m−1
.
25x−3 × 89−2x
= 1 for x.
4x
4 Solve the following inequations.
a 2 × 5x + 5x < 75
3
576
WE2
.
Solve
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
b
1
9
2x−3
>
1
9
7−x
Solve 30 × 102x + 17 × 10x − 2 = 0 for x.
6 Solve 2x − 48 × 2−x = 13 for x.
7 WE4 a Express each of the following numbers in scientific notation, and state the
number of significant figures each number contains.
i 1 409 000
ii 0.000 130 6
b Express the following as basic numerals.
i 3.04 × 105
ii 5.803 × 10−2
8 Calculate (4 × 106) 2 × (5 × 10−3) without using a calculator.
9 a Express the following in index (exponent) form.
Consolidate
Apply the most
appropriate
mathematical
processes and tools
WE3
a5
×
b−4
b Express the following in surd form.
a3 b 4
i
ii
3
1
3
a2 b
O
FS
5
5
i a2 ÷ b2
ii22
iii3
O
10 Evaluate without a calculator:
3
b 3−1 + 50 − 22 × 9
c 23 ×
4
9
1
−2
PR
a 42
÷ (6 × (3−2) 2)
d
15 × 5
−2
5
−1
2
3
2
1
1
c
3(x2y−2) 3
b
G
(3x4y2) −1
2
(2mn−2) −2
m−1n
3(m2n)
d
3
2
32 × 2 × (ab) 2
(−8a2) 2b2
f
4m2n−2 × −2 m2n2
(−3m3n−2) 2
4x − 1 − 2x(4x − 1)
−1
2
O
N
LI
N
E
m−1 − n−1
m2 − n2
32 × 43x
12 a Express
as a power of 2.
16x
31+n × 81n−2
b Express
as a power of 3.
243n
e
1
×
3 2
10n4m−1
÷
2a3b−3
3a3b−1
PA
a
E
1252 − 202
11 Simplify and express the answer with positive indices.
c Express 0.001 ×
d Express
−
4
5n+1
3
5n
5
2
10 × 100 × (0.1)
−2
3
as a power of 10.
as a power of 5.
13 Solve for x:
a 22x × 82−x × 16
c 9x ÷ 271−x =
e 45x + 45x =
− 3x
2
3
8
24x−5
=
2
4x
b 253x−3 ≤ 1254+x
d
2
3
2x
3−2x
>
27
8
−1
3
×
1
214
3x
f 5 3 × 5 2 = 25x+4
Topic 11 Exponential functions 577
14 Use a suitable substitution to solve the following equations.
a 32x − 10 × 3x + 9 = 0
c 25x + 52+x − 150 = 0
e 10x − 102−x = 99
b 24 × 22x + 61 × 2x = 23
d (2x + 2−x) 2 = 4
f 23x + 3 × 22x−1 − 2x = 0
15 a Express in scientific notation:
i−0.000 000 050 6
iithe diameter of the Earth, given its radius is 6370 km
iii3.2 × 104 × 5 × 10−2
ivthe distance between Roland Garros and Kooyong
ii 1.44 × 106
O
FS
tennis stadiums of 16 878.7 km.
b Express as a basic numeral:
i6.3 × 10−4 + 6.3 × 104
1
2
PR
O
c Express the following to 2 significant figures:
i60 589 people attended a football match.
iiThe probability of winning a competition is 1.994 × 10−2.
iiiThe solution to an equation is x = −0.006 34.
ivThe distance flown per year by the Royal Flying Doctor Service is
26 597 696 km.
16 If x = 3 + 3
−1
3
, show that x3 − 3x = 10
.
3
E
1
3
17 a Solve the pair of simultaneous equations for x and y:
G
1
52x−y = 125
PA
102y−6x = 0.01
b Solve the pair of simultaneous equations for a and k.
a × 2k−1 = 40
O
n−1
÷
3x
a
n+1
=
3
x
, determine the values of the constants a and n.
4
19 Evaluate:
N
Master
2x2
3a
LI
N
18 If
E
a × 22k−2 = 10
a 5−4.3 and express the answer in scientific notation to 4 significant figures
b 22.9 ÷ 1.3E2 to 4 significant figures and explain what the 1.3E2 notation
means
c 5.04 × 10−6 ÷ (3 × 109), expressing the answer in standard form
d 5.04 × 10−6 ÷ (3.2 × 104.2), expressing the answer in standard form to
4 significant figures.
x2y−2
20 a Simplify 1
.
2x3 y5
b Solve the equations:
i5x × 252x =
1
5
to obtain x exactly
ii5x × 252x = 0.25 to obtain x to 4 significant figures.
578 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
11.3
Units 1 & 2
AOS 1
Topic 8
Concept 2
Not all solutions to indicial equations are rational. In order to obtain the solution to an
equation such as 2x = 5, we need to learn about logarithms.
Index-logarithm forms
A logarithm is also another name for an index.
The index statement, n = ax, with base a and index x, can be expressed with the index
as the subject. This is called the logarithm statement and is written as x = loga (n).
The statement is read as ‘x equals the log to base a of n’ (adopting the abbreviation of
‘log’ for logarithm).
O
FS
Indices as
logarithms
Concept summary
Practice questions
Indices as logarithms
The statements n = ax and x = loga (n) are equivalent.
• n = ax ⇔ x = loga (n), where the base a ∈ R+ \ 1 , the
number n ∈ R + and the logarithm, or index, x ∈ R.
O
Consider again the equation 2x = 5. The solution is obtained by converting this index
statement to the logarithm statement.
PR
2x = 5
∴ x = log2 (5)
PA
G
E
The number log2 (5) is irrational: the power of 2 which gives the number 5 is
not rational. A decimal approximation for this logarithm can be obtained using
a calculator. The exact solution to the indicial equation 2x = 5 is x = log2 (5); an
approximate solution is x ≈ 2.3219 to 5 significant figures.
Not all expressions containing logarithms are irrational. Solving the equation 2x = 8
by converting to logarithm form gives:
2x = 8
∴ x = log2 (8)
E
In this case, the expression log2 (8) can be simplified. As the power of 2 which gives
the number 8 is 3, the solution to the equation is x = 3; that is, log2 (8) = 3.
O
N
LI
N
Use of a calculator
Calculators have two inbuilt logarithmic functions.
Base 10 logarithms are obtained from the LOG key. Thus log10 (2) is evaluated as
log(2), giving the value of 0.3010 to 4 decimal places. Base 10 logarithms are called
common logarithms and, in a previous era, they were commonly used to perform
calculations from tables of values. They are also known as Briggsian logarithms in
deference to the English mathematician Henry Briggs who first published their table
of values in the seventeenth century.
Base e logarithms are obtained from the LN key. Thus loge (2) is evaluated as ln (2),
giving the value 0.6931 to 4 decimal places. Base e logarithms are called natural or
Naperian logarithms, after their inventor John Napier, a seventeenth-century Scottish
baron with mathematical interests. These logarithms occur extensively in calculus.
The number e itself is known as Euler’s number and, like π, it is a transcendental
irrational number that has great importance in higher mathematical studies, as you
will begin to discover in Units 3 and 4 of Mathematical Methods.
CAS technology enables logarithms to bases other than 10 or e to be evaluated.
Topic 11 Exponential functions 579
WORKeD
exaMpLe
5
a Express 34 = 81 as a logarithm statement.
1
= −2 as an index statement.
49
c Solve the equation 10x = 12.8, expressing the exponent x to 2 significant figures.
d Solve the equation log5 (x) = 2 for x.
b Express log7
tHinK
WritE
a 1 Identify the given form.
34 = 81
The index form is given with base 3, index or
logarithm 4 and number 81.
Since n = ax ⇔ x = loga (n),
81 = 34 ⇒ 4 = log3 (81).
The logarithm statement is 4 = log3 (81).
1
b log7
= −2
49
The logarithm form is given with base 7,
1
number
and logarithm, or index, of −2.
49
Since x = loga (n) ⇔ n = ax
1 ⇒ 1
−2 = log7
= 7−2
49
49
1
The index statement is
= 7−2.
49
c 10x = 12.8
∴ x = log10 (12.8)
2 Convert to the equivalent form.
E
PA
c 1 Convert to the equivalent form.
G
2 Convert to the equivalent form.
PR
O
b 1 Identify the given form.
O
FS
a
2 Evaluate using a calculator and state the
E
answer to the required accuracy.
Note: The base is 10 so use the LOG
key on the calculator.
LI
N
d 1 Convert to the equivalent form.
2 Calculate the answer.
≈ 1.1 to 2 significant figures
d log5 (x) = 2
∴
x = 52
= 25
O
N
Logarithm laws
Since logarithms are indices, unsurprisingly there is a set of laws which control
simplification of logarithmic expressions with the same base.
For any a, m, n > 0, a ≠ 1, the laws are:
1. loga (1) = 0
2. loga (a) = 1
3. loga (m) + loga (n) = loga (mn)
m
4. loga (m) − loga (n) = loga
n
p
5. loga (m ) = ploga (m).
Note that there is no logarithm law for either the product or quotient of logarithms or
for expressions such as loga (m ± n).
580
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
Proofs of the logarithm laws
1. Consider the index statement.
a0 = 1
∴ loga (1) = 0
2. Consider the index statement.
a1 = a
∴ loga (a) = 1
3. Let x = loga (m) and y = loga (n).
Convert to logarithm form:
O
x + y = loga (mn)
O
FS
∴ m = ax and n = ay
mn = ax × ay
= ax+y
PR
Substitute back for x and y:
loga (m) + loga (n) = loga (mn)
4. With x and y as given in law 3:
G
E
m ax
=
n ay
= ax−y
PA
Converting to logarithm form, and then substituting back for x and y gives:
x − y = loga
m
n
LI
N
E
Substitute back for x and y:
∴ loga (m) − loga (n) = loga
m
n
5. With x as given in law 3:
O
N
x = loga (m)
∴ m = ax
Raise both sides to the power p:
∴ (m) p = (ax) p
∴ mp = apx
Express as a logarithm statement with base a:
px = loga (mp)
Substitute back for x:
∴ p loga (m) = loga (mp)
Topic 11 Exponential functions 581
WORKeD
exaMpLe
6
Simplify the following using the logarithm laws.
a log10 (5) + log10 (2)
c log3 (2a4) + 2 log3
b log2 (80) − log2 (5)
a
2
loga
d
1
4
loga (2)
tHinK
WritE
a 1 Apply the appropriate logarithm law.
a log10 (5) + log10 (2) = log10 (5 × 2)
= log10 (10)
= 1 since loga (a) = 1
O
FS
2 Simplify and state the answer.
80
5
= log2 (16)
b log2 (80) − log2 (5) = log2
b 1 Apply the appropriate logarithm law.
= log2 (24)
= 4log2 (2)
O
2 Further simplify the logarithmic expression.
PR
3 Calculate the answer.
E
G
PA
E
2 Apply an appropriate logarithm law to combine
LI
N
the two terms as one logarithmic expression.
N
3 State the answer.
O
d 1 Express the numbers in the logarithm terms in
index form.
Note: There is no law for division of logarithms.
2 Simplify the numerator and denominator
separately.
3 Cancel the common factor in the numerator and
denominator.
582
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
a
2
c log3 (2a4) + 2log3
c 1 Apply a logarithm law to the second term.
Note: As with indices, there is often more
than one way to approach the simplification of
logarithms.
=4×1
=4
a
2
1
2
= log2 (2a4) + 2 × 12loga
a
2
= log2 (2a4) + 2loga
= log2 (2a4) + loga
= log3 2a4 ×
a
2
a
2
= log3 (a5)
= 5log3 (a)
1
loga (2−2)
4
d
=
loga (2)
loga (2)
loga
=
=
−2loga (2)
loga (2)
−2loga (2)
loga (2)
= −2
Logarithms as operators
O
FS
Just as both sides of an equation may be raised to a power and the equality still holds,
taking logarithms of both sides of an equation maintains the equality.
If m = n, then it is true that loga (m) = loga (n) and vice versa, provided the same base
is used for the logarithms of each side.
This application of logarithms can provide an important tool when solving indicial
equations.
Consider again the equation 2x = 5 where the solution was given as x = log2 (5).
Take base 10 logarithms of both sides of this equation.
2x = 5
log10 (2x) = log10 (5)
It also demonstrates that log2 (5) =
PR
O
Using one of the logarithm laws, this becomes x log10 (2) = log10 (5) from which the
log10 (5)
solution to the indicial equation is obtained as x =
. This form of the solution
log10 (2)
can be evaluated on a scientific calculator and is the prime reason for choosing base 10
logarithms in solving the indicial equation.
log10 (5)
E
, which is a particular example of another
log10 (2)
logarithm law called the change of base law.
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G
Change of base law for calculator use
The equation ax = p for which x = loga (p) could be solved in a similar way to 2x = 5,
log10 (p)
log10 (p)
giving the solution as x =
. Thus loga (p) =
. This form enables
log10 (a)
log10 (a)
decimal approximations to logarithms to be calculated on scientific calculators.
The change of base law is the more general statement allowing base a logarithms to
logb (p)
. This more general
be expressed in terms of any other base b as loga (p) =
logb (a)
form shall be left until Units 3 and 4.
WORKeD
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7
Convention
There is a convention that if the base of a logarithm is not stated, this implies it is
base 10. As it is on a calculator, log(n) represents log10 (n). When working with
base 10 logarithms it can be convenient to adopt this convention.
a State the exact solution to 5x = 8 and calculate its value to 3 decimal places.
b Calculate the exact value and the value to 3 decimal places of the solution
to the equation 21−x = 6x.
tHinK
WritE
a 1 Convert to the equivalent form and state the
a 5x = 8
exact solution.
∴ x = log5 (8)
The exact solution is x = log5 (8).
topic 11 expOnentIaL FunCtIOns
583
2 Use the change of base law to express the answer
Since
in terms of base 10 logarithms.
loga (p) =
then
log5 (8) =
∴
x=
log10 (p)
log10 (a)
log10 (8)
log10 (5)
log10 (8)
log10 (5)
b 21−x = 6x
b 1 Take base 10 logarithms of both sides.
O
FS
∴ x ≈ 1.292 to 3 decimal places.
3 Calculate the approximate value.
Take logarithms to base 10 of both sides:
log(21−x) = log(6x)
Note: The convention is not to write the base 10.
(1 − x) log(2) = x log(6)
3 Solve the linear equation in x.
4 Calculate the approximate value.
G
E
Note: This is no different to solving any other
linear equation of the form a − bx = cx except the
constants a, b, c are expressed as logarithms.
Expand:
log(2) − x log(2) = x log(6)
Collect x terms together:
log(2) = x log(6) + x log(2)
= x(log(6) + log(2))
log(2)
x=
log(6) + log(2)
This is the exact solution.
O
longer exponents.
PR
2 Apply the logarithm law so that x terms are no
x ≈ 0.279 to 3 decimal places.
PA
Note: Remember to place brackets around the
denominator for the division.
E
equations containing logarithms
O
N
LI
N
While the emphasis in this topic is on exponential (indicial) relations for which
some knowledge of logarithms is essential, it is important to know that logarithms
contribute substantially to Mathematics. As such, some equations involving logarithms
are included, allowing further consolidation of the laws which logarithms must satisfy.
Remembering the requirement that x must be positive for loga (x) to be real, it is advisable
to check any solution to an equation involving logarithms. Any value of x which when
substituted back into the original equation creates a ‘loga (negative number)’ term
must be rejected as a solution. Otherwise, normal algebraic approaches together with
logarithm laws are the techniques for solving such equations.
WORKeD
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8
Solve the equation log6 (x) + log6 (x − 1) = 1 for x.
tHinK
1 Apply the logarithm law which reduces the
equation to one logarithm term.
584
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
WritE
log6 (x) + log6 (x − 1) = 1
∴ log6 (x(x − 1)) = 1
∴ log6 (x2 − x) = 1
2 Convert the logarithm form to its
equivalent form.
Note: An alternative method is to write
log6 (x2 − x) = log6 (6) from which x2 − x = 6
is obtained.
Converting from logarithm form to index
form gives:
x2 − x = 61
∴ x2 − x = 6
x2 − x − 6 = 0
3 Solve the quadratic equation.
∴ (x − 3)(x + 2) = 0
∴ x = 3, x = −2
Check in log6 (x) + log6 (x − 1) = 1
O
FS
4 Check the validity of both solutions in the
original equation.
If x = 3, LHS = log6 (3) + log6 (2)
= log6 (6)
=1
O
= RHS
PR
If x = −2, LHS = log6 (−2) + log6 (−3) which
is not admissible.
Therefore reject x = −2.
The solution is x = 3.
E
5 State the answer.
Exercise 11.3 Indices as logarithms
a Express 54 = 625 as a logarithm statement.
G
WE5
1
as an index statement.
2
c Solve the equation 10x = 8.52, expressing the exponent x to 2
significant figures.
d Solve the equation log3 (x) = −1 for x.
2 a Evaluate loge (5) to 4 significant figures and write the equivalent index
statement.
b Evaluate 103.5 to 4 significant figures and write the equivalent logarithm
statement.
3 WE6 Simplify using the logarithm laws:
a log12 (3) + log12 (4)
b log2 (192) − log2 (12)
b Express log36 (6) =
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E
Work without CAS
Q1–4
1
PA
PRactise
c log3
(3a3)
− 2log3 a
3
2
d
loga (8)
loga (4)
4 Given loga (2) = 0.3 and loga (5) = 0.7, evaluate:
a loga (0.5)
b loga (2.5)
c loga (20)
x
5 WE7 a State the exact solution to 7 = 15 and calculate its value to 3 decimal places.
b Calculate the exact value and the value to 3 decimal places of the solution to the
equation 32x+5 = 4x.
6 If log2 (3) − log2 (2) = log2 (x) + log2 (5), solve for x.
7 WE8 Solve the equation log3 (x) + log3 (2x + 1) = 1 for x.
8 Solve the equation log6 (x) − log6 (x − 1) = 2 for x.
Topic 11 Exponential functions 585
Consolidate
3
iii10−3 = 0.001
ii42 = 8
i 25 = 32
b Express as an index statement:
1
iiilog10 (0.1) = −1
2
10 Rewrite each of the following in the equivalent index or logarithm form and hence
calculate the value of x.
1
a x = log2
8
b log25 (x) = −0.5
c 10(2x) = 4. Express the answer to 2 decimal places.
d 3 = e−x. Express the answer to 2 decimal places.
e logx (125) = 3
f logx (25) = −2
11 Use the logarithm laws to evaluate the following.
a log9 (3) + log9 (27)
b log9 (3) − log9 (27)
c 2log2 (4) + log2 (6) − log2 (12)
d log5 ( log3 (3))
i log2 (16) = 4
iilog9 (3) =
e log11
7
7
1
11
+ 2log11
− log11
− log11
3
3
3
9
5
PA
log a ( 3)
c log8 (16) × log16 (8)
G
x×x
x2
12 Simplify the following.
log a (9)
E
3
2
f log2
a
PR
O
O
FS
Apply the most
appropriate
mathematical
processes and tools
9 a Express as a logarithm statement with the index as the subject:
13 a Express log2 (10) in terms of log10 (2).
b 1 − log4 (3n + 3n+1)
d log10 (1000) + log0.1
1
1000
E
b State the exact solution and then give the approximate solution to 4 significant
O
N
LI
N
figures for each of the following indicial equations.
i 11x = 18
ii5−x = 8
iii72x = 3
c Obtain the approximate solution to 4 significant figures for each of the
following inequations.
i 3x ≤ 10
ii5−x > 0.4
d Solve the following equations.
i 2log5(x) = 8
ii2log2(x) = 7
14 Solve the indicial equations to obtain the value of x to 2 decimal places.
a 71−2x = 4
b 10−x = 5x−1
c 52x−9 = 37−x
d 103x+5 = 62−3x
e 0.254x = 0.82−0.5x
f 4x+1 × 31−x = 5x
15 Solve the following for x.
a log2 (2x + 1) + log2 (2x − 1) = 3log2 (3)
b log3 (2x) + log3 (4) = log3 (x + 12) − log3 (2)
c log2 (2x + 12) − log2 (3x) = 4
d log2 (x) + log2 (2 − 2x) = −1
e (log10 (x) + 3)(2log4 (x) − 3) = 0
f 2log3 (x) − 1 = log3 (2x − 3)
586 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
16 Given loga (3) = p and loga (5) = q, express the following in terms of p and q.
a loga (15)
b loga (125)
c loga (45)
d loga (0.6)
25
81
17 Express y in terms of x.
a log10 (y) = log10 (x) + 2
e loga
c 2log2
y
f loga (
b log2 (x2
= 6x − 2
2
5) × loga ( 27)
y) = x
d x = 10y−2
f 103log10(y) = xy
O
FS
e log10 (103xy) = 3
a 22x − 14 × 2x + 45 = 0
b 5−x − 5x = 4
c 92x − 31+2x + 2 = 0
d loga (x3) + loga (x2) − 4loga (2) = loga (x)
f
log10 (x + 1)
= log10 (x)
19 a Give the solution to 12x = 50 to 4 significant figures.
b Give the exact solution to the equation log(5x) + log(x + 5) = 1.
E
Master
log10 (x3)
PR
e (log2 (x)) 2 − log2 (x2) = 8
O
18 Solve the following equations, giving exact solutions.
G
20 a Evaluate log10 (5) + log5 (10) with the calculator on ‘Standard’ mode and
Graphs of exponential functions
E
Exponential functions are functions of the form f : R → R, f(x) = ax, a ∈ R + \ 1 .
They provide mathematical models of exponential growth and exponential decay
situations such as population increase and radioactive decay respectively.
LI
N
11.4
PA
explain the answer obtained.
b Evaluate logy (x) × logx (y) and explain how the result is obtained.
Units 1 & 2
AOS 1
N
Topic 8
Concept 3
O
Graphs of
exponential
functions
Concept summary
Practice questions
Interactivity
Exponential functions
int-5959
The graph of y = ax where a > 1
Before sketching such a graph, consider the table of
values for the function with rule y = 2x.
x
y = 2x
−3 −2 −1
1
8
1
4
1
2
0
1
2
3
1
2
4
8
From the table it is evident that 2x > 0 for all values of x,
and that as x → −∞, 2x → 0. This means that the graph
will have a horizontal asymptote with equation y = 0.
It is also evident that as x → ∞, 2x → ∞ with the values
increasing rapidly.
y
y = 2x
(1, 2)
(0, 1)
0
y=0
x
Topic 11 Exponential functions 587
Since these observations are true for any function y = ax
where a > 1, the graph of y = 2x will be typical of the basic
graph of any exponential with base larger than 1.
Key features of the graph of y = 2x and any such function
y = ax where a > 1:
• horizontal asymptote with equation y = 0
• y-intercept is (0, 1)
• shape is of ‘exponential growth’
• domain R
• range R +
• one-to-one correspondence
y = 10x y = 2x
y
(1, 10)
(1, 2)
(0, 1)
O
FS
0
y=0
x
The graph of y = a x where 0 < a < 1
O
For y = 2x, the graph contains the point (1, 2); for the
graph of y = ax, a > 1, the graph contains the point (1, a), showing that as the base
increases, the graph becomes steeper more quickly for values x > 0. This is illustrated
by the graphs of y = 2x and y = 10x, with the larger base giving the steeper graph.
E
PR
y
An example of a function whose rule is in the form y = ax
x
x
y = 2–x
1
1
where 0 < a < 1 is y =
. Since
= 2−x, the rule
2
2
x
1
for the graph of this exponential function y =
where
2
the base lies between 0 and 1 is identical to the rule
1
(–1, 2)
1, –
−x
2
y = 2 where the base is greater than 1.
(0, 1)
The graph of y = 2−x shown is typical of the
0 y=0
graph of y = a−x where a > 1 and of the graph of
y = ax where 0 < a < 1.
Key features of the graph of y = 2−x and any such function
with rule expressed as either y = ax where 0 < a < 1 or as y = a−x where a > 1:
x
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G
( )
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• horizontal asymptote with equation y = 0
• y-intercept is (0, 1)
• shape is of ‘exponential decay’
• domain R
• range R +
• one-to-one correspondence
• reflection of y = 2x in the y-axis
The basic shape of an exponential function is either one of ‘growth’ or ‘decay’.
y
y
(0, 1)
(0, 1)
(0, 1)
0
y
y=0
x
y = ax, a > 1
0
y = a–x, a > 1
y=0
x
0
y=0
x
y = ax, 0 < a < 1
As with other functions, the graph of y = −ax will be inverted (reflected in the x-axis).
588 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
WORKeD
exaMpLe
9
a On the same set of axes, sketch the graphs
of y = 5x and y = −5x, stating their ranges.
b Give a possible equation for the graph
shown.
tHinK
WritE
a 1 Identify the asymptote of the
a y = 5x
second point.
4 Use the relationship between the
y
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E
5 Sketch and label each graph.
6 State the range of each graph.
b 1 Use the shape of the graph to
suggest a possible form for the rule.
2 Use a given point on the graph to
calculate a.
3 State the equation of the graph.
O
y = −5x
This is the reflection of y = 5x in the x-axis.
The graph of y = −5x has the same asymptote as that
of y = 5x.
Equation of its asymptote is y = 0.
Its y-intercept is (0, −1).
Point (1, −5) lies on the graph.
PA
two functions to deduce the key
features of the second function.
Let x = 1.
y = 51
=5
⇒ (1, 5)
PR
3 Calculate the coordinates of a
E
2 Find the y-intercept.
The asymptote is the line with equation y = 0.
y-intercept: when x = 0, y = 1 ⇒ (0, 1)
G
first function.
O
FS
y=0
(1, 5)
y = 5x
(0, 1)
y=0
x
0
(0, –1)
y = –5x
(1, –5)
The range of y = 5x is R + and the range of y = −5x is R−.
b The graph has a ‘decay’ shape.
Let the equation be y = a−x.
The point (−1, 5) ⇒ 5 = a1
∴
a=5
The equation of the graph could be y = 5−x.
1
The equation could also be expressed as y =
5
y = 0.2x.
x
or
topic 11 expOnentIaL FunCtIOns
589
translations of exponential graphs
Once the basic exponential growth or exponential decay shapes are known, the graphs
of exponential functions can be translated in similar ways to graphs of any other
functions previously studied.
y
y = 3x+1
y = 3x
y = 3x – 1
(0, 3)
(1, 3)
(1, 2)
(–1, 1)
(0, 1)
x
0
y = –1
10
Sketch the graphs of each of the following and state the range of each.
a y = 2x − 4
PA
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PR
O
the graph of y = a x − h
Under a horizontal translation the asymptote is
unaffected. The point on the y-axis will no longer
occur at y = 1. An additional point to the y-intercept
that can be helpful to locate is the one where x = h,
since ax−h will equal 1 when x = h.
A horizontal translation and a vertical translation of
the graph of y = 3x are illustrated in the diagram by
the graphs of y = 3x+1 and y = 3x − 1 respectively.
Under the horizontal translation of 1 unit to the
left, the point (0, 1) → (−1, 1); under the vertical
translation of 1 unit down, the point (1, 3) → (1, 2).
O
FS
the graph of y = a x + k
Under a vertical translation the position of the asymptote will be altered to y = k.
If k < 0, the graph will have x-axis intercepts which are found by solving the
exponential equation ax + k = 0.
tHinK
WritE
a 1 State the equation of the
a y = 2x − 4
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2 Calculate the y-intercept.
3 Calculate the x-intercept.
O
The vertical translation 4 units down affects the asymptote.
The asymptote has the equation y = −4.
E
asymptote.
y-intercept: let x = 0,
y=1−4
= −3
y-intercept is (0, −3).
x-intercept: let y = 0,
2x − 4 = 0
∴ 2x = 4
∴ 2x = 22
∴x=2
x-intercept is (2, 0).
590
b y = 10−(x+1)
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
4 Sketch the graph and
A ‘growth’ shape is expected since the coefficient of x is positive.
state the range.
y
(2, 0)
y = 2x – 4
x
0
y = –4
Range is (−4, ∞).
Reflection in y-axis, horizontal translation 1 unit to the left. The
asymptote will not be affected.
Asymptote: y = 0
There is no x-intercept.
y-intercept: let x = 0,
y = 10−1
PR
from the given equation.
b y = 10−(x+1)
O
b 1 Identify the key features
O
FS
(0, –3)
1
10
y-intercept is (0, 0.1).
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N
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state the range.
G
E
3 Sketch the graph and
Let x = −1
y = 100
PA
2 Calculate the coordinates
of a second point on
the graph.
E
=
=1
The point (−1, 1) lies on the graph.
A ‘decay’ shape is expected since the coefficient of x is negative.
y
y = 10–(x+1)
(–1, 1)
(0, 0.1)
0
y=0 x
Range is R + .
Dilations
Exponential functions of the form y = b × ax have been dilated by a factor
b (b > 0) from the x-axis. This affects the y-intercept, but the asymptote
remains at y = 0.
Topic 11 Exponential functions 591
Combinations of transformations
y
y = 3x
y = 32x
(1, 6)
y = 2 × 3x
(0, 2)
(1, 3)
1, 3
–
2
(0, 1)
( )
0
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FS
Exponential functions of the form y = anx have been
1
dilated by a factor (n > 0) from the y-axis. This
n
affects the steepness of the graph but does not affect
either the y-intercept or the asymptote.
A dilation from the x-axis of factor 2 and a dilation
1
from the y-axis of factor of the graph of y = 3x are
2
illustrated in the diagram by the graphs of y = 2 × 3x
and y = 32x respectively. Under the dilation from
the x-axis of factor 2, the point (1, 3) → (1, 6);
1
under the dilation from the y-axis of factor ,
2
1
the point (1, 3) → , 3 .
2
x
11
Sketch the graphs of each of the following and state the range of each.
PA
WORKeD
exaMpLe
G
E
PR
O
Exponential functions with equations of the form y = b × an(x−h) + k are derived
from the basic graph of y = ax by applying a combination of transformations. The key
features to identify in order to sketch the graphs of such exponential functions are:
• the asymptote
• the y-intercept
• the x-intercept, if there is one.
Another point that can be obtained simply could provide assurance about the shape.
Always aim to show at least two points on the graph.
b y = 1 − 3 × 2 −x
a y = 10 × 52x−1
tHinK
WritE
O
N
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N
the given equation.
E
a 1 Identify the key features using
2 Calculate the coordinates of a
second point.
a y = 10 × 52x−1
asymptote: y = 0
no x-intercept
y-intercept: let x = 0
y = 10 × 5−1
1
= 10 ×
5
=2
y-intercept is (0, 2).
Since the horizontal translation is
y = 10 × 5
1
2× 2 −1
= 10 × 50
= 10 × 1
= 10
Point
592
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
1
, 10 lies on the graph.
2
1
1
to the right, let x = .
2
2
y
3 Sketch the graph and state
(0.5, 10)
the range.
y = 10 × 52x−1
(0, 2)
y=0
O
FS
0
Range is R + .
y = b × an(x−h) + k and state
the asymptote.
b y = 1 − 3 × 2−x
∴ y = −3 × 2−x + 1
Asymptote: y = 1
y-intercept: let x = 0
y = −3 × 20 + 1
PR
2 Calculate the y-intercept.
O
b 1 Write the equation in the form
= −3 × 1 + 1
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E
PA
Note: As the point (0, −2)
lies below the asymptote and
the graph must approach the
asymptote, there will be an
x-intercept.
x-intercept: let y = 0
0 = 1 − 3 × 2−x
1
2−x =
3
In logarithm form,
1
−x = log2
3
= log2 (3−1)
= −log2 (3)
∴ x = log2 (3)
The exact x-intercept is (log2 (3), 0).
G
3 Calculate the x-intercept.
E
= −2
y-intercept is (0, −2).
4 Calculate an approximate
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N
value for the x-intercept to
help determine its position on
the graph.
x = log2 (3)
=
log10 (3)
log10 (2)
≈ 1.58
The x-intercept is approximately (1.58, 0).
Topic 11 Exponential functions 593
y
5 Sketch the graph and state
the range.
Note: Label the x-intercept with
its exact coordinates once the
graph is drawn.
y=1
0
x
(log2(3), 0)
(0, –2)
y=1–3×
2x
O
FS
Range is (−∞, 1).
Exercise 11.4 Graphs of exponential functions
WE9
a On the same set of axes, sketch
the graphs of y = 3 and y = −3 , stating
their ranges.
b Give a possible equation for the
graph shown.
2
2 Sketch the graphs of y = (1.5) x and y =
3
on the same set of axes.
x
x
PR
Work without CAS
1
O
PRactise
y
(–1, 3)
x
E
(0, 1)
0
y=0 x
Sketch the graphs of each of the
following and state the range of each.
a y = 4x − 2
b y = 3−(x+2)
4 Sketch the graph of y = 4x−2 + 1 and state its range.
5 WE11 Sketch the graphs of each of the following and state the range of each.
1
a y = × 101−2x
b y = 5 − 4 × 3−x
2
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PA
3
6 The graph shown has the equation y = a.3x + b. Determine the values of a and b.
y
O
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y=2
594 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
(0, 0)
0
x
Consolidate
x
x
x
1
1
1
b i Sketch, on the same set of axes, the graphs of y =
,y =
and y =
.
4
6
8
ii Express each rule in a different form.
8a i Sketch, on the same set of axes, the graphs of y = 5−x, y = 7−x and y = 9−x.
ii Describe the effect produced by increasing the base.
bi Sketch, on the same set of axes, the graphs of y = (0.8) x, y = (1.25) x
and y = (0.8) −x.
ii Describe the relationships between the three graphs.
9 Sketch each of the following graphs, showing the asymptote and labelling any
intersections with the coordinate axes with their exact coordinates.
a y = 5−x + 1
b y = 1 − 4x
x
c y = 10 − 2
d y = 6.25 − (2.5) −x
10 Sketch the graphs of:
a y = 2x−2
b y = −3x+2
c y = 4x−0.5
d y = 71−x
11 Sketch the graphs of:
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Apply the most
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7 ai Sketch, on the same set of axes, the graphs of y = 4x, y = 6x and y = 8x.
ii Describe the effect produced by increasing the base.
a y = 3 × 2x
3x
b y = 2 4
c y = −3 × 2−3x
d y = 1.5 × 10
−
x
2
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12 a Sketch the graphs of y = 32x and y = 9x and explain the result.
b i Use index laws to obtain another form of the rule for y = 2 × 40.5x.
ii Hence or otherwise, sketch the graph of y = 2 × 40.5x.
y
(0, 5)
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13 a
Determine a possible rule for the given graph in the form y = a.10x + b.
0
y=3
x
b The graph of an exponential function of the form y = a.3kx contains the points (1, 36) and (0, 4). Determine its rule and state the equation of its asymptote.
y
c For the graph shown,
y=6
determine a possible rule in
the form y = a − 2 × 3b−x.
d Express the equation given
in part c in another form
not involving a horizontal
translation.
(0, 0)
x
Topic 11 Exponential functions 595
14 Sketch the graphs of the following exponential functions and state their ranges.
Where appropriate, any intersections with the coordinate axes should be given
to 1 decimal place.
x
2
a y = 2 × 102x − 20
b y = 5 × 21−x − 1
c y = 3 − 2
3
x+1
2x−1
d y = 2(3.5)
e y = 8 − 4 × 5
f y = −2 × 103x−1 − 4
−7
15 Consider the function f : R → R, f(x) = 3 − 6 × 2
x−1
2
.
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a Evaluate:
i f(1)
ii f(0), expressing the answer in simplest surd form.
b For what value of x, if any, does:
i f(x) = −9
ii f(x) = 0
iii f(x) = 9?
c Sketch the graph of y = f(x) and state its range.
d Solve the inequation f(x) ≥ −1 correct to 2 significant figures.
16 Use a graphical means to determine the number of intersections between:
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a y = 2x and y = −x, specifying an interval in which the x-coordinate of any point
Master
1
2
x
× 162, giving the coordinates of any points of intersection.
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f y = 22x−1 and y =
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of intersection lies.
b y = 2x and y = x2
c y = ex and y = 2x
d y = 2−x + 1 and y = sin (x)
e y = 3 × 2x and y = 6x, determining the coordinates of any points of intersection
algebraically.
17 Obtain the coordinates of the points of intersection of y = 2x and y = x2.
18 Sketch the graphs of y1 = 33 − 2(11) x and y2 = 33 − 2(11) x+1 and compare their
Applications of exponential functions
The importance of exponential functions lies in the frequency with which they occur
in models of phenomena involving growth and decay situations, in chemical and
physical laws of nature and in higher-level mathematical analysis.
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asymptotes, x- and y-intercepts and the value of their x-coordinates when
y = 10. What transformation maps y1 to y2?
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Units 1 & 2
AOS 1
Topic 8
Concept 4
Applications of
exponential
functions
Concept summary
Practice questions
596 Exponential growth and decay models
For time t, the exponential function defined by y = b × ant where a > 1 represents
exponential growth over time if n > 0 and exponential decay over time if
n < 0. The domain of this function would be restricted according to the way
the independent time variable t is defined. The rule y = b × ant may also be
written as y = b.ant.
In some mathematical models such as population growth, the initial population may be
represented by a symbol such as N0. For an exponential decay model, the time it takes
for 50% of the initial amount of the substance to decay is called its half-life.
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
WORKeD
exaMpLe
12
The decay of a radioactive substance is modelled by Q(t) = Q0 × 2.7−kt where
Q kg is the amount of the substance present at time t years and Q0 and k are
positive constants.
a Show that the constant Q0 represents the initial amount of the substance.
b If the half-life of the radioactive substance is 100 years, calculate k to one
significant figure.
c If initially there was 25 kg of the radioactive substance, how many kilograms
would decay in 10 years? Use the value of k from part b in the calculations.
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a 1 Calculate the initial amount.
a Q(t) = Q0 × 2.7−kt
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The initial amount is the value of Q when t = 0.
Let t = 0:
= Q0
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Q(0) = Q0 × 2.70
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Therefore Q0 represents the initial amount of
the substance.
initial amount of the substance to decay.
Since the half-life is 100 years, when t = 100,
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information.
Note: It does not matter that the value of
Q0 is unknown since the Q0 terms cancel.
b The half-life is the time it takes for 50% of the
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b 1 Form an equation in k from the given
2 Solve the exponential equation to obtain k
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to the required accuracy.
c 1 Use the values of the constants to state the
actual rule for the exponential decay model.
2 Calculate the amount of the substance
present at the time given.
Q(100) = 50% of Q0
Q(100) = 0.50Q0 ..... (1)
From the equation, Q(t) = Q0 × 2.7−kt.
When t = 100, Q(100) = Q0 × 2.7−k(100)
∴ Q(100) = Q0 × 2.7−100k .... (2)
Equate equations (1) and (2):
0.50Q0 = Q0 × 2.7−100k
Cancel Q0 from each side:
0.50 = 2.7−100k
Convert to the equivalent logarithm form.
−100k = log2.7 (0.5)
1
k=−
log (0.5)
100 2.7
log10 (0.5)
1
=−
×
100 log10 (2.7)
≈ 0.007
c Q0 = 25, k = 0.007
∴ Q(t) = 25 × 2.7−0.007t
When t = 10,
Q(10) = 25 × 2.7−0.07
≈ 23.32
topic 11 expOnentIaL FunCtIOns
597
Since 25 − 23.32 = 1.68, in 10 years
approximately 1.68 kg will have decayed.
3 Calculate the amount that has decayed.
Note: Using a greater accuracy for the
value of k would give a slightly different
answer for the amount decayed.
analysing data
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One method for detecting if data has an exponential relationship can be carried out using
logarithms. If the data is suspected of following an exponential rule such as y = A × 10kx,
then the graph of log(y) against x should be linear. The reasoning for this is as follows:
y = A × 10kx
y
∴ = 10kx
A
y
∴ log
= kx
A
∴ log(y) − log(A) = kx
∴ log(y) = kx + log(A)
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This equation can be written in the form Y = kx + c where Y = log(y) and c = log(A).
The graph of Y versus x is a straight line with gradient k and vertical axis Y-intercept
(0, log(A)).
Such an analysis is called a semi-log plot. While experimental data is unlikely to give
a perfect fit, the equation would describe the line of best fit for the data.
Logarithms can also be effective in determining a power law that connects variables.
If the law connecting the variables is of the form y = x p then log(y) = p log(x).
Plotting log(y) values against log(x) values will give a straight line of gradient p if the
data does follow such a law. Such an analysis is called a log-log plot.
For a set of data {(x, y)}, plotting log(y) versus
log(x) gave the straight line shown in the
diagram.
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exaMpLe
log(y)
Form the equation of the graph and
hence determine the rule connecting y
and x.
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(0, 3)
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(2, 0)
tHinK
1 State the gradient and the coordinates of the
intercept with the vertical axis.
0
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Gradient =
−3
2
Intercept with vertical axis: (0, 3)
=
598
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
rise
run
log(x)
3 Express the equation in terms of the
variables marked on the axes of the
given graph.
4 Collect the terms involving logarithms
together and simplify to create a logarithm
statement.
Let Y = log(y) and X = log(x).
The equation of the line is Y = mX + c where
3
m = − , c = 3.
2
3
Therefore the equation of the line is Y = − X + 3.
2
The vertical axis is log(y) and the horizontal
axis is log(x), so the equation of the graph is
3
log(y) = − log(x) + 3.
2
∴ log(y) = −1.5 log(x) + 3
log(y) + 1.5 log (x) = 3
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2 Form the equation of the line.
log(y) + log(x1.5) = 3
log(yx1.5) = 3
∴ log10 (yx1.5) = 3
5 Express the equation with y as the subject.
yx1.5 = 103
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Note: Remember the base of the
logarithm is 10.
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y = 1000x−1.5
Exercise 11.5 Applications of exponential functions
The decay of a radioactive substance is modelled by Q(t) = Q0 × 1.7−kt
where Q is the amount of the substance present at time t years and Q0 and k are
positive constants.
a Show that the constant Q0 represents the initial amount of the substance.
b If the half-life of the radioactive substance is 300 years, calculate k to one
significant figure.
c If initially there was 250 kg of the radioactive substance, how many kilograms
would decay in 10 years? Use the value of k from part b in the calculations.
2 The manager of a small business is concerned about the amount of time she
spends dealing with the growing number of emails she receives. The manager
starts keeping records and finds the average number of emails received per day
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1
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PRactise
t
can be modelled by D = 42 × 216 where D is the average number of emails
received per day t weeks from the start of the records.
a How many daily emails on average was the manager receiving when she
log(y)
commenced her records?
b After how many weeks does the
model predict that the average
number of emails received per day
(0, 2)
will double?
3 WE13 For a set of data (x, y) ,
plotting log(y) versus log(x) gave the
(–0.8, 0)
straight line shown in the diagram.
0
log(x)
From the equation of the graph
and hence determine the rule
connecting y and x.
Topic 11 Exponential functions 599
4 For a set of data (x, y) , the semi-log plot of
Consolidate
log(y) versus x gave the straight line shown in
the diagram.
Form the equation of the graph and hence
determine an exponential rule connecting
(1, 0.3)
x
y and x.
0 (0, 0)
5 The value V of a new car depreciates so that its
value after t years is given by V = V0 × 2−kt.
a If 50% of the purchase value is lost in 5 years,
calculate k.
b How long does it take for the car to lose 75% of its
purchase value?
6 The number of drosophilae (fruit flies), N, in a
colony after t days of observation is modelled by
N = 30 × 20.072t. Give whole-number answers to the
following.
a How many drosophilae were present when the colony was initially observed?
b How many of the insects were present after 5 days?
c How many days does it take the population number to double from its initial value?
d Sketch a graph of N versus t to show how the population changes.
e After how many days will the population first exceed 100?
7 The value of an investment which earns compound interest can be calculated from
nt
r
the formula A = P 1 +
where P is the initial investment, r the interest rate
n
per annum (yearly), n the number of times per year the interest is compounded
and t the number of years of the investment.
An investor deposits $2000 in an account where interest is compounded monthly.
a If the interest rate is 3% per annum:
iShow that the formula giving the value of the investment is
A = 2000(1.0025) 12t.
iiCalculate how much the investment is worth after a 6-month period.
iiiWhat time period would be needed for the value of the investment to
reach $2500?
b The investor would like the $2000 to grow to $2500 in a shorter time period.
What would the interest rate, still compounded monthly, need to be for the goal
to be achieved in 4 years?
8 A cup of coffee is left to cool on a kitchen table inside a
Brisbane home. The temperature of the coffee T (°C) after
t minutes is thought to be given by T = 85 × 3−0.008t.
a By how many degrees does the coffee cool in 10 minutes?
b How long does it take for the coffee to cool to 65 °C?
c Sketch a graph of the temperature of the
coffee for t ∈ [0, 40].
d By considering the temperature the model predicts the coffee will eventually
cool to, explain why the model is not realistic in the long term.
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Apply the most
appropriate
mathematical
processes and tools
log(y)
600 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
9 The contents of a meat pie immediately after being heated in a microwave have a
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temperature of 95 °C. The pie is removed from the microwave and left to cool.
A model for the temperature of the pie as it cools is given by T = a × 3−0.13t + 25
where T is the temperature after t minutes of cooling.
a Calculate the value of a.
b What is the temperature of the contents of the pie after
being left to cool for 2 minutes?
c Determine how long, to the nearest minute, it will take for
the contents of the meat pie to cool to 65 °C.
d Sketch the graph showing the temperature over time and
state the temperature to which this model predicts the
contents of the pie will eventually cool if left unattended.
10 The barometric pressure P, measured in kilopascals, at height h above sea level,
measured in kilometres, is given by P = Po × 10−kh where Po and k are positive
constants. The pressure at the top of Mount Everest is approximately one third
that of the pressure at sea level.
a Given the height of Mount Everest is approximately 8848 metres, calculate the
value of k to 2 significant figures.
Use the value obtained for k for the remainder of question 6.
b Mount Kilimanjaro has a height of approximately 5895 metres. If the
atmospheric pressure at its summit is approximately 48.68 kilopascals, calculate
the value of P0 to 3 decimal places.
c Use the model to estimate the atmospheric pressure to 2 decimal places at the
summit of Mont Blanc, 4810 metres, and of Mount Kosciuszko, 2228 metres
in height.
d Draw a graph of the atmospheric pressure against height showing the readings
for the four mountains from the above information.
11 The common Indian mynah bird was introduced into Australia in
order to control insects affecting market gardens in Melbourne.
It is now considered to be Australia’s most important pest
problem. In 1976, the species was introduced to an urban area
in New South Wales. By 1991 the area averaged 15 birds per
square kilometre and by 1994 the density reached an average of
75 birds per square kilometre.
Topic 11 Exponential functions 601
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A model for the increasing density of the mynah bird population is thought to be
D = D0 × 10kt where D is the average density of the bird per square kilometre
t years after 1976 and D0 and k are constants.
a Use the given information to set up a pair of simultaneous equations
in D and t.
1
b Solve these equations to show that k = log(5) and D0 = 3 × 5−4 and hence
3
that k ≈ 0.233 and D0 ≈ 0.005.
c A project was introduced in 1996 to curb the growth in numbers of these
birds. What does the model predict was the average density of the mynah
bird population at the time the project was introduced in the year 1996? Use
k ≈ 0.233 and D0 ≈ 0.005 and round the answer to the nearest whole number.
d Sometime after the project is successfully implemented, a different model for
the average density of the bird population becomes applicable. This model is
t
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given by D = 30 × 10 3 + b. Four years later, the average density is reduced
to 40 birds per square kilometre. How much can the average density expect to
be reduced?
12 Carbon dating enables estimates of the age of fossils of once living organisms
to be ascertained by comparing the amount of the radioactive isotope carbon-14
remaining in the fossil with the normal amount present in the living entity, which
can be assumed to remain constant during the organism’s life. It is known that
carbon-14 decays with a half-life of approximately 5730 years according to an
kt
1
exponential model of the form C = Co ×
, where C is the amount of the
2
isotope remaining in the fossil t years after death and Co is the normal amount of
the isotope that would have been present when the organism was alive.
a Calculate the exact value of the positive constant k.
b The bones of an animal are unearthed during digging explorations by a mining
company. The bones are found to contain 83% of the normal amount of the
isotope carbon-14. Estimate how old the bones are.
13 a
Obtain the equation of the given linear graphs and hence determine the
relationship between y and x.
log(y)
i
iThe linear graph of log10 (y) against
log10 (x) is shown.
(2, 0)
iiThe linear graph of log2 (y) against
log(x)
x is shown.
(0,
–1)
b The acidity of a solution is due
to the presence of hydrogen ions.
The concentration of these ions is
measured by the pH scale calculated as
pH = − log ([H + ]) where [H + ] is the
log2(y)
ii
concentration of hydrogen ions.
iThe concentration of hydrogen ions
(0, 0)
in bleach is 10−13 per mole and in
x
pure water the concentration is 10−7
(4, –1)
per mole. What are the pH readings for
bleach and for pure water?
602 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
iiLemon juice has a pH reading of 2 and milk has a
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pH reading of 6. Use scientific notation to express
the concentration of hydrogen ions in each of lemon
juice and milk and then write these concentrations
as numerals.
iiiSolutions with pH smaller than 7 are acidic and
those with pH greater than 7 are alkaline. Pure
water is neutral. How much more acidic is lemon
juice than milk?
ivFor each one unit of change in pH, explain the
effect on the concentration of hydrogen ions and
acidity of a solution.
14 The data shown in the table gives the population of
Australia, in millions, in years since 1960.
1975
1990
x (years since 1960)
30
15
y (population in millions)
13.9
17.1
log(y)
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53
22.9
a Complete the third row of the table by
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evaluating the log(y) values to
2 decimal places.
b Plot log(y) against x and construct a
straight line to fit the points.
c Show that the equation of
the line is approximately
Y = 0.006x + 1.05 where Y = log(y).
d Use the equation of the line to show that
the exponential rule between y and x is
approximately y = 11.22 × 100.006x.
e After how many years did the population double the 1960 population?
f It is said that the population of Australia is likely to exceed 28 million by the
year 2030. Does this model support this claim?
15 Experimental data yielded the following table of values:
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x
1
1.5
2
2.5
y
5.519
6.483
7.615
8.994
3
3.5
4
10.506 12.341 14.496
a Enter the data as lists in the Statistics menu and obtain the rule connecting the
data by selecting the following from the Calc menu:
iExponential Reg
iiLogarithmic Reg
b Graph the data on the calculator to confirm which rule better fits the data.
16 Following a fall from his bike, Stephan is feeling some shock but not, initially,
a great deal of pain. However, his doctor gives him an injection for relief from
the pain that he will start to feel once the shock of the accident wears off. The
amount of pain Stephan feels over the next 10 minutes is modelled by the function
Topic 11 Exponential functions 603
Topic 8
Concept 5
Inverses of
exponential
functions
Concept summary
Practice questions
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The exponential function has a one-to-one correspondence so its inverse must also be
a function. To form the inverse of y = ax, interchange the x- and y-coordinates.
function: y = ax domain R, range R +
inverse function: x = ay domain R + , range R
∴ y = loga (x)
Therefore, the inverse of an exponential function is a logarithmic function: y = loga (x)
and y = ax are the rules for a pair of inverse functions. These are transcendental
functions, not algebraic functions. However, they can be treated similarly to the
inverse pairs of algebraic functions previously encountered. This means the graph of
y = loga (x) can be obtained by reflecting the graph of y = ax in the line y = x.
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AOS 1
The inverse of y = a x, a ∈ R +\{1}
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Units 1 & 2
Inverses of exponential functions
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P(t) = (200t + 16) × 2.7−t, where P is the measure of pain on a scale from 0 to
100 that Stephan feels t minutes after receiving the injection.
a Give the measure of pain Stephan is feeling:
iat the time the injection is administered
ii15 seconds later when his shock is wearing off but the injection has not
reached its full effect.
b Use technology to draw the graph showing Stephan’s pain level over the
10-minute interval and hence give, to 2 decimal places:
ithe maximum measure of pain he feels
iithe number of seconds it takes for the injection to start lowering his
pain level
iiihis pain levels after 5 minutes and after 10 minutes have elapsed.
c Over the 10-minute interval, when was the effectiveness of the
injection greatest?
d At the end of the 10 minutes, Stephan receives a second injection
modelled by P(t) = (100(t − 10) + a) × 2.7−(t−10), 10 ≤ t ≤ 20.
iDetermine the value of a.
iiSketch the pain measure over the time interval t ∈ [0, 20] and label the
maximum points with their coordinates.
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The graph of y = loga (x),
for a > 1
604 The shape of the basic logarithmic
graph with rule y = loga (x), a > 1
is shown as the reflection in the line
y = x of the exponential graph with
rule y = ax, a > 1.
The key features of the graph of
y = loga (x) can be deduced from
those of the exponential graph.
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
y
y = ax
(1, a)
y=x
y = logax
(0, 1)
(a, 1)
(1, 0)
x
y = ax
y = loga(x)
horizontal asymptote with equation y = 0
x-intercept (1, 0)
point (1, a) lies on the graph
range R +
domain R
one-to-one correspondence
vertical asymptote with equation x = 0
y-intercept (0, 1)
point (a, 1) lies on the graph
domain R +
range R
one-to-one correspondence
Note that logarithmic growth is much slower than exponential growth and also note
> 0, if x > 1
= 0, if x = 1
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that, unlike ax which is always positive, loga (x)
< 0, if 0 < x < 1
14
a Form the exponential rule for the inverse of y = log2 (x) and hence deduce
the graph of y = log2 (x) from the graph of the exponential.
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The logarithmic function is formally written as f : R + → R, f(x) = loga (x).
b Given the points (1, 2), (2, 4) and (3, 8) lie on the exponential graph in
part a, explain how these points can be used to illustrate the logarithm law
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log2 (m) + log2 (n) = log2 (mn).
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a 1 Form the rule for the inverse by
a y = log2 (x)
Inverse: x = log2 (y)
∴ y = 2x
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interchanging coordinates and then
make y the subject of the rule.
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tHinK
y = 2x
Asymptote: y = 0
y-intercept: (0, 1)
second point: let x = 1,
∴y=2
Point (1, 2) is on the graph.
y
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y = 2x
(1, 2)
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2 Sketch the exponential function.
(0, 1)
0
y=0
x
topic 11 expOnentIaL FunCtIOns
605
y = log2 (x) has:
asymptote: x = 0
x-intercept: (1, 0)
second point: (2, 1)
3 Reflect the exponential graph
in the line y = x to form the
required graph.
y
y=x
(1, 2)
y = log2(x)
(0, 1)
y = 2x
(1, 0)
x
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0
(2, 1)
b 1 State the coordinates of the
b Given the points (1, 2), (2, 4) and (3, 8) lie on the
exponential graph, the points (2, 1), (4, 2) and (8, 3) lie
on the graph of y = log2 (x).
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corresponding points on the
logarithm graph.
each of the points on the
logarithmic graph.
3 Use the relationship between the
The sum of the y-coordinates of the points on y = log2 (x)
when x = 2 and x = 4 equals the y-coordinate of the
point on y = log2 (x) when x = 8, as 1 + 2 = 3.
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y-coordinates to illustrate the
logarithm law.
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y = log2 (x)
point (2, 1): when x = 2, y = 1
point (4, 2): when x = 4, y = 2
point (8, 3): when x = 8, y = 3
2 State the x- and y-values for
log2 (2) + log2 (4) = log2 (8)
PA
log2 (2) + log2 (4) = log2 (2 × 4)
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This illustrates the logarithm law
log2 (m) + log2 (n) = log2 (mn)
with m = 2 and n = 4.
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The inverse of exponential functions of the form
y = b × an(x − h) + k
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The rule for the inverse of y = b × an(x−h) + k is calculated in the usual way by
interchanging x- and y-coordinates to give x = b × an(y−h) + k.
Expressing this equation as an index statement:
x − k = b × an(y−h)
x−k
= an(y−h)
b
Converting to the equivalent logarithm form:
loga
606 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
x−k
= n(y − h)
b
Rearranging to make y the subject:
x−k
1
loga
n
b
=y−h
y=
x−k
1
loga
+h
n
b
15
Consider the function f : R → R, f (x) = 5 × 2 −x + 3.
a What is the domain of its inverse?
PR
WORKeD
exaMpLe
O
O
FS
The inverse of any exponential function is a logarithmic function. Since the
corresponding pair of graphs of these functions must be symmetric about the line
y = x, this provides one approach for sketching the graph of any logarithmic function.
In this section, the graphs of logarithmic functions are obtained by deduction using
the previously studied key features of exponential functions under a sequence of
transformations. Should either the exponential or the logarithmic graph intersect the
line y = x then the other graph must also intersect that line at exactly the same point.
Due to the transcendental nature of these functions, technology is usually required to
obtain the coordinates of any such point of intersection.
b Form the rule for the inverse function and express the inverse function as a
mapping.
E
c Sketch y = f (x) and y = f −1 (x) on the same set of axes.
WritE
a 1 Determine the range
a f(x) = 5 × 2−x + 3
the inverse.
b 1 Form the rule for
N
LI
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the inverse function
by interchanging
x- and y-coordinates
and rearranging the
equation obtained.
Note: Remember to
express the rule for
the inverse function
f −1 with f −1 (x)
in place of y as
its subject.
O
The domain of the inverse function is the range of the given function.
The domain of the inverse function is (3, ∞).
E
2 State the domain of
Asymptote: y = 3
y-intercept: (0, 8)
This point lies above the asymptote, so the range of f is (3, ∞).
PA
of the given function.
G
tHinK
b Let f(x) = y.
Function: y = 5 × 2−x + 3
Inverse:
x = 5 × 2−y + 3
∴ x − 3 = 5 × 2−y
∴ x − 3 = 2−y
5
Converting to logarithm form:
x−3
−y = log
2
5
x−3
∴ y = −log2
5
The inverse function has the rule f −1 (x) = −log2
x−3
.
5
topic 11 expOnentIaL FunCtIOns
607
2 Write the inverse
function as
a mapping.
The inverse function has domain (3, ∞) and its rule is
x−3
f −1 (x) = −log2
.
5
Hence, as a mapping:
f −1 : (3, ∞) → R, f −1 (x) = −log2
of the exponential
function f and use
this to deduce the
graph of the inverse
function f −1.
c f(x) = 5 × 2−x + 3, f −1 (x) = −log2
x−3
5
The key features of f determine the key features of f −1.
y = f(x) y = f −1 (x)
asymptote: y = 3 asymptote: x = 3
y-intercept (0, 8) x-intercept (8, 0)
second point on y = f(x):
let x = −1
O
FS
c 1 Sketch the graph
x−3
5
= 13
PR
O
y = 5 × 21 + 3
Point (−1, 13) is on y = f(x) and the point (13, −1) is on y = f −1 (x).
y
x=3
G
(0, 8)
y=3
(8, 0)
x
0
(13, –1)
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y = 5 × 2–x + 3
y=x
E
(–1, 13)
x –3
y = –log2 – —
5
The graphs intersect on the line y = x.
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N
Relationships between the inverse pairs
As the exponential and logarithmic functions are a pair of inverses, each ‘undoes’ the
effect of the other. From this it follows that:
loga (ax) = x and aloga(x) = x
The first of these statements could also be explained using logarithm laws:
loga (ax) = x loga (a)
=x×1
=x
608 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
The second statement can also be explained from the index-logarithm definition that
an = x ⇔ n = loga (x). Replacing n by its logarithm form in the definition gives:
an = x
a loga(x) = x
WORKeD
exaMpLe
16
a Simplify log12 (22x × 3x) using the inverse relationship between exponentials
and logarithms.
tHinK
WritE
a 1 Use index laws to simplify the product of
a log12 (22x × 3x)
O
FS
b Evaluate 102 log10(5).
Consider the product 22x × 3x.
22x × 3x = (22) x × 3x
powers given in the logarithm expression.
= 4x × 3 x
O
= (4 × 3) x
PR
= 12x
log12 (22x × 3x) = log12 (12x)
2 Simplify the given expression.
b 102 log10(5)
b 1 Apply a logarithm law to the term in
the index.
E
= 10 log10(5)
PA
G
2 Simplify the expression.
=x
2
= 10 log10(25)
= 25
Therefore, 102 log10(5) = 25.
E
transformations of logarithmic graphs
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Knowledge of the transformations of graphs enables the graph of any logarithmic
function to be obtained from the basic graph of y = loga (x). This provides an
alternative to sketching the graph as the inverse of that of an exponential function.
Further, given the logarithmic graph, the exponential graph could be obtained as the
inverse of the logarithmic graph.
The logarithmic graph under a combination of transformations will be studied in
Units 3 and 4. In this section we shall consider the effect a single transformation has
on the key features of the graph of y = loga (x).
Dilations
Dilations from either coordinate axis are recognisable from the equation of the
x
would give
logarithmic function: for example, y = 2 loga (x) and y = loga
2
the images when y = loga (x) undergoes a dilation of factor 2 from the x-axis and
from the y-axis respectively. The asymptote at x = 0 would be unaffected by either
dilation. The position of the x-intercept is affected by the dilation from the y-axis as
(1, 0) → (2, 0). The dilation from the x-axis does not affect the x-intercept.
topic 11 expOnentIaL FunCtIOns
609
E
PA
G
E
PR
O
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Horizontal translation: the graph of y = loga (x − h)
The vertical asymptote will always be affected by a horizontal translation and this
affects the domain of the logarithmic function. Under a horizontal translation of h units
to the right or left, the vertical asymptote at x = 0 must move h units to the right or
left respectively. Hence, horizontally translating the graph of y = loga (x) by h units to
obtain the graph of y = loga (x − h) produces the following changes to the key features:
• equation of asymptote: x = 0 → x = h
• domain: x : x > 0 → x : x > h
• x-intercept: (1, 0) → (1 + h, 0)
These changes are illustrated
y = log2(x)
y
x=3
in the diagram by the graph
x=0
of y = log2 (x) and its image,
y = log2 (x − 3), after a horizontal
y = log2(x – 3)
translation of 3 units to the right.
(2, 1)
The diagram shows that the
(5, 1)
domain of y = log2 (x − 3) is
(3, ∞). Its range is unaffected
x
(1, 0)
0
(4, 0)
by the horizontal translation and
remains R.
It is important to realise that
the domain and the asymptote
position can be calculated
algebraically, since we only take
logarithms of positive numbers.
For example, the domain of
y = log2 (x − 3) can be calculated by solving the inequation x − 3 > 0 ⇒ x > 3. This
means that the domain is (3, ∞) as the diagram shows. The equation of the asymptote
of y = log2 (x − 3) can be calculated from the equation x − 3 = 0 ⇒ x = 3.
The function defined by y = loga (nx + c) would have a vertical asymptote when
nx + c = 0 and its domain can be calculated by solving nx + c > 0.
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Vertical translation: the graph of y = loga (x) + k
Under a vertical translation of k units, neither the
y
y = log2(x)
domain nor the position of the asymptote alters
x=0
from that of y = loga (x). The translated graph
(2, 1)
will have an x-intercept which can be obtained by
(1,
0)
0
solving the equation loga (x) + k = 0.
(2, –2)
The graph of y = log2 (x) − 3 is a vertical
translation down by 3 units of the graph of
y = log2(x) – 3
y = log2 (x). Solving log2 (x) − 3 = 0 gives x = 23
so the graph cuts the x-axis at x = 8, as illustrated.
(8, 0)
Reflections: the graphs of y = −loga (x) and y = loga (−x)
The graph of y = −loga (x) is obtained by inverting the graph of y = loga (x); that is,
by reflecting it in the x-axis.
The graph of y = loga (−x) is obtained by reflecting the graph of y = loga (x) in the
y-axis. For loga (−x) to be defined, −x > 0 so the graph has domain x : x < 0 .
610 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
x
y
x=0
y = loga(–x)
(–1, 0)
0
y = loga(x)
x
(1, 0)
O
FS
y = –loga(x)
17
a Sketch the graph of y = log2 (x + 2) and state its domain.
b Sketch the graph of y = log10 (x) + 1 and state its domain.
PR
WORKeD
exaMpLe
O
The relative positions of the graphs of y = log a (x), y = −log a (x) and y = log a (−x)
are illustrated in the diagram. The vertical asymptote at x = 0 is unaffected by either
reflection.
c The graph of the function for which
y
f(x) = log2 (b − x) is shown below.
E
i Determine the value of b.
ii State the domain and range of,
x=2
(0, 1)
0
(1, 0)
x
PA
G
and form the rule for, the inverse
function.
iii Sketch the graph of y = f −1 (x).
tHinK
E
a 1 Identify the transformation involved.
LI
N
2 Use the transformation to state
the equation of the asymptote and
the domain.
N
3 Calculate any intercepts with the
O
coordinate axes.
Note: The domain indicates there will
be an intercept with the y-axis as well as
the x-axis.
WritE
a y = log2 (x + 2)
Horizontal translation 2 units to the left
The vertical line x = 0 → the vertical line
x = −2 under the horizontal translation.
The domain is x : x > −2 .
y-intercept: when x = 0,
y = log2 (2)
=1
y-intercept (0, 1)
x-intercept: when y = 0,
log2 (x + 2) = 0
x + 2 = 20
x+2=1
x = −1
x-intercept (−1, 0)
Check: the point (1, 0) → (−1, 0) under the
horizontal translation.
topic 11 expOnentIaL FunCtIOns
611
x = –2
4 Sketch the graph.
y
y = log2 (x + 2)
(0, 1)
(–1, 0)
b 1 Identify the transformation involved.
x
0
b y = log10 (x) + 1
O
FS
Vertical translation of 1 unit upwards
2 State the equation of the asymptote and
The vertical transformation does not affect either
the position of the asymptote or the domain.
Hence, the equation of the asymptote is x = 0.
The domain is R + .
the domain.
Since the domain is R + there is no y-intercept.
x-intercept: when y = 0,
O
3 Obtain any intercept with the
coordinate axes.
4 Calculate the coordinates of a second
LI
N
E
PA
point on the graph.
G
E
PR
log10 (x) + 1 = 0
log10 (x) = −1
x = 10−1
1
=
or 0.1
10
x-intercept is (0.1, 0).
=0+1
=1
The point (1, 1) lies on the graph.
Check: the point (1, 0) → (1, 1) under the
vertical translation.
y x=0
y = log10 (x) + 1
O
N
5 Sketch the graph.
Point: let x = 1.
y = log 10 (1) + 1
612 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
(1, 1)
0
(0.1, 0)
x
c i 1 State the equation of the asymptote
shown in the graph and use this to
calculate the value of b.
Note: The function rule can be
rearranged to show the horizontal
translation and a reflection in
the y-axis.
f(x) = log2 (b − x)
c i f(x) = log2 (b − x)
From the diagram, the asymptote of the graph
is x = 2.
From the function rule, the asymptote occurs
when:
b−x=0
x=b
Hence, b = 2.
= log2 (−(x − b))
ii 1 Give the domain and range of the
O
FS
The horizontal translation determines
the position of the asymptote.
ii The given function has domain (−∞, 2) and
range R. Therefore the inverse function has
domain R and range (−∞, 2).
O
inverse function.
function f : y = log2 (2 − x)
2 Form the rule for the inverse function
by interchanging x- and y-coordinates
and rearranging the equation obtained.
PR
inverse f −1 : x = log2 (2 − y)
2x = 2 − y
E
∴ y = 2 − 2x
∴ f −1 (x) = 2 − 2x
iii The key features of f give those for f −1.
iii 1 Use the features of the logarithm
PA
G
graph to deduce the features of the
exponential graph.
y
y=2
(0, 1)
y = f –1(x)
(1, 0)
0
x
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N
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N
E
2 Sketch the graph of y = f −1 (x).
y = f(x)
y = f −1 (x)
asymptote: x = 2asymptote y = 2
x-intercept (1, 0)
y-intercept (0, 1)
y-intercept (0, 1)
x-intercept (1, 0)
Topic 11 Exponential functions 613
Exercise 11.6 Inverses of exponential functions
1
WE14
a Form the exponential rule for the inverse of y = log10 (x) and hence
deduce the graph of y = log10 (x) from the graph of the exponential.
b Given the points (1, 10), (2, 100) and (3, 1000) lie on the exponential
graph in part a, explain how these points can be used to illustrate the
m
logarithm law log10 (m) − log10 (n) = log10
.
n
2 Sketch the graphs of y = 2−x and its inverse and form the rule for this inverse.
3 WE15 Consider the function f : R → R, f(x) = 4 − 23x.
a What is the domain of its inverse?
b Form the rule for the inverse function and express the inverse function as
a mapping.
c Sketch y = f(x) and y = f −1 (x) on the same set of axes.
4 Consider the function defined by y = 4 + 2log2 (x).
a Form the rule for the inverse function.
b Sketch the graph of the inverse function and hence draw the graph of
y = 4 + 2log2 (x) on the same set of axes.
c In how many places do the two graphs intersect?
5 WE16 a Simplify log6 (22x × 9x) using the inverse relationship between
exponentials and logarithms.
b Evaluate 2−3 log2(10).
6 Simplify 5x log5(2) − log5(3).
7 WE17 a Sketch the graph of y = log10 (x − 1) and state its domain.
b Sketch the graph of y = log5 (x) − 1 and state
x = –b
y
its domain.
c The graph of the function for which
f(x) = − log2 (x + b) is shown.
(0, 0)
iDetermine the value of b.
x
0
iiState the domain and range of, and form the
rule for, the inverse function.
y = –log2 (x + b)
iiiSketch the graph of y = f −1 (x).
8 Sketch the graph of the function f : R + → R, f(x) = 1 − log4 (x) by identifying the
transformations involved.
9 a On the same axes, sketch the graphs of y = 3x and y = 5x together with
their inverses.
b State the rules for the inverses graphed in part a as logarithmic functions.
c Describe the effect of increasing the base on a logarithm graph.
d On a new set of axes, sketch y = 3x + 1 and draw its inverse.
e Give the equation of the inverse of y = 3x + 1.
f Sketch the graphs of y = 5x+1 and its inverse on the same set of axes and give
the rule for the inverse.
N
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E
PA
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E
PR
O
O
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PRactise
Consolidate
O
Apply the most
appropriate
mathematical
processes and tools
614 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
10 Sketch the graphs of the following functions and their inverses and form the rule
for the inverse.
a y = 2
−x
3
b y =
d y = 3x+1 + 3
1
2
× 8x − 1
c y = 2 − 4x
f y = 21−x
e y = −10−2x
11 Given f : R → R, f(x) = 8 − 2 × 32x:
a Determine the domain and the range of the inverse function f −1.
b Evaluate f(0).
c At what point would the graph of f −1 cut the x-axis?
d Obtain the rule for f −1 and express f −1 as a mapping.
O
FS
12 Consider the function defined by y = 2 × (1.5) 2−x.
O
a For what value of x does y = 2?
b For what value of y does x = 0?
c Sketch the graph of y = 2 × (1.5) 2−x showing the key features.
d On the same set of axes sketch the graph of the inverse function.
e Form the rule for the inverse.
PR
f Hence state the solution to the equation 2 × (1.5) 2−x = 2 − log1.5
13 Given g: (−1, ∞) → R, g(x) = −3log5 (x + 1) :
x
.
2
PA
G
E
a State the range of g−1.
b Evaluate g(0).
c At what point would the graph of g−1 cut the x-axis?
d Obtain the rule for g−1 and express g−1 as a mapping.
e Sketch the graph of y = g−1 (x).
f Use the graph in part e to deduce the graph of y = g(x).
14 a Evaluate:
i 3log3(8)
ii 10log10(2) +log10(3)
iii5−log5(2)
1 log (25)
6
iv 62
b Simplify:
E
i 3log3(x)
6x+1 − 6x
5
15 The diagram shows the graph of the exponential function y = 2ax+b + c. The
graph intersects the line y = x twice and cuts the x-axis at 12, 0 and the
y-axis at (0, −2).
y = 2ax+b + c
y
a Form the rule for the exponential function.
b Form the rule for the inverse function.
y=x
c For the inverse, state the equation of its asymptote
1 ,0
–
and the coordinates of the points where its graph
2
would cut the x- and y-axes.
x
0
d Copy the diagram and sketch the graph of the inverse
(0, –2)
on the same diagram. How many points of intersection
y = –4
of the inverse and the exponential graphs are there?
e The point (log2 (3), k) lies on the exponential graph.
Calculate the exact value of k.
f Using the equation for the inverse function, verify that the point (k, log2 (3)) lies
on the inverse graph.
N
LI
N
iiilog2 (2x) + log3 (9x)
O
ii 23log2(x)
iv log6
( (
Topic 11 Exponential functions 615
16 Sketch the graphs of the following transformations of the graph of y = loga (x),
O
N
LI
N
E
PA
G
E
PR
O
O
FS
stating the domain and range, equation of the asymptote and any points of
intersection with the coordinate axes.
a y = log5 (x) − 2
b y = log5 (x − 2)
c y = log10 (x) + 1
d y = log3 (x + 1)
e y = log3 (4 − x)
f y = −log2 (x + 4)
17 a Describe the transformations which map y = log2 (x) → y = −2log2 (2 − x).
b Use a logarithm law to describe the vertical translation
which maps y = log2 (x) → y = log2 (2x).
c Use the change of base law to express y = log2 (x) in terms of base 10
logarithms and hence describe the dilation which
would map y = log10 (x) → y = log2 (x).
d Hick’s Law arose from research into the time taken for a
person to make a decision when faced with a number of
possible choices. For n equally probable options, the law is
expressed as t = blog2 (n + 1) where t is the time taken to
choose an option, b is a positive constant and n ≥ 2. Draw a
sketch of the time against the number of choices and show
that doubling the number of options does not double the
time to make the choice between them.
18 a The graph of the function with equation y = alog7 (bx) contains the points (2, 0)
and (14, 14). Determine its equation.
b The graph of the function with equation y = alog3 (x) + b contains the
1
points , 8 and (1, 4).
3
y
iDetermine its equation.
x = –2
iiObtain the coordinates of the point
where the graph of the inverse
function would cut the y-axis.
(–1.5, 0)
c ­ iFor the graph illustrated
x
0
in the diagram, determine
a possible equation in
(0, –2)
the form y = alog2 (x − b) + c.
iiUse the diagram to sketch the
graph of the inverse and form the
rule for the inverse.
d Consider the functions f and g for
which f(x) = log3 (4x + 9) and g(x) = log4 (2 − 0.1x).
iDetermine the maximal domain of each function.
iiState the equations of the asymptotes of the graphs of y = f(x) and y = g(x).
iiiCalculate the coordinates of the points of intersection of each of the graphs
with the coordinate axes.
ivSketch the graphs of y = f(x) and y = g(x) on separate diagrams.
616 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
2−x
2
with its inverse. Express the values to 4 significant figures, where appropriate.
20 Consider the two functions with rules y = log2 (x + 4) and y = log2 (x) + log2 (4).
a iShould the graphs of y = log2 (x + 4) and y = log2 (x) + log2 (4) be the same
graphs? Use CAS technology to sketch the graphs of y = log2 (x + 4) and
y = log2 (x) + log2 (4) to verify your answer.
ii
Give any values of x for which the graphs have the same value and justify
algebraically.
b Sketch the graph of y = 2log3 (x), stating its domain, range and type of
correspondence.
c Sketch the graph of y = log3 (x2), stating its domain, range and type of
correspondence.
d The graphs in parts b and c are not identical. Explain why this does not
contradict the logarithm law loga (mp) = ploga (m).
O
N
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E
PA
G
E
PR
O
O
FS
Master
19 Obtain the coordinates of any points of intersection of the graph of y = 2 × 3
Topic 11 Exponential functions 617
11.7 Review
the Maths Quest review is available in a customisable
format for you to demonstrate your knowledge of this
topic.
• Extended-response questions — providing you with
the opportunity to practise exam-style questions.
a summary of the key points covered in this topic is
also available as a digital document.
REVIEW QUESTIONS
Download the Review questions document from
the links found in the Resources section of your
eBookPLUS.
ONLINE ONLY
PR
O
the review contains:
• short-answer questions — providing you with the
opportunity to demonstrate the skills you have
developed to efficiently answer questions without the
use of CAS technology
• Multiple-choice questions — providing you with the
opportunity to practise answering questions using
CAS technology
www.jacplus.com.au
O
FS
ONLINE ONLY
Activities
G
www.jacplus.com.au
PA
Interactivities
Units 1 & 2
Exponential functions
O
N
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N
E
A comprehensive set of relevant interactivities
to bring difficult mathematical concepts to life
can be found in the Resources section of your
eBookPLUS.
studyON is an interactive and highly visual online
tool that helps you to clearly identify strengths
and weaknesses prior to your exams. You can
then confidently target areas of greatest need,
enabling you to achieve your best results.
E
to access eBookPlUs activities, log on to
618
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
Sit topic test
11 Answers
12a 25+2x
Exercise 11.2
b 3−7
c 103
d 5n
1a 29n
1
2 2
b 6a b
8
9
−16pq8
13a x = 1
c
3 x = 8
4a x < 2
5 x = −1
10
3
O
b x <
6 x = 4
PR
7a i 1.409 × 106; 4 significant figures
ii 1.306 × 10−4; 4 significant figures
b i 304 000
ii 0.058 03
E
8 8 × 1010
3
G
9a i a2b2
19 7
ii a 6 b3
5
1
9
E
ii
PA
a
b3
32
b i iii
b 0
c 162
d 25
N
9x10
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N
10a 8
y4
3
O
11a
b
b 0.1762; 1.3E2 = 1.3 × 102
20a
−8n5
5
x3
2y2
y5
1
=
x3
9
2y2
b i −5
9
−1
mn(m + n)
2x − 1
(4x − 1)
19a 9.873 × 10−4
5
3m n
c
40
f
;n = 6
d 9.938 × 10−11
1
3 2
e
−11
2
18a = ± 36 × 2
c 1.68 × 10−15
5
3 2
16a b
d
7
10
1
d x >
2
1
e x =
2
f x = 48
14a x = 0, 2
b x = −3
c x = 1
d x = 0
e x = 2
f x = −1
15a i −5.06 × 10−8
ii 1.274 × 104 km
iii 1.6 × 103
iv 1.687 87 × 104 km
b i 63 000.000 63
ii 1200
c i 61 000
ii 0.020
iii x = −0.0063
iv 27 million km
16Proof required — check with your teacher
17a x = 4, y = 11
b a = 160, k = −1
c x =
m4
O
FS
2
b x ≤ 6
1
2
ii −0.1723
Exercise 11.3
1a 4 = log5 (625)
1
b 6 = 362
Topic 11 Exponential functions 619
c x ≈ 0.93
c i x ≤ 2.096
1
d x =
3
2a loge (5) ≈ 1.609; 5 = e1.609
b 103.5 ≈ 3162; 3.5 = log10 (3162)
3a 1
b 4
c 1
3
d
2
4a −0.3
b 0.4
c 1.3
5a x = log7 (15); x ≈ 1.392
5log(3)
b x =
; x = −6.774
log(4) − 2log(3)
3
6 x =
10
7 x = 1
36
8 x =
35
9a i 5 = log2 (32)
3
2
d i 125
ii 7
14a 0.14
b 0.41
c 5.14
d −0.65
e 0.08
b
c
d
e
f
16a
b
= log4 (8)
c
iii −3 = log10 (0.001)
1
2
d
= 16
e
E
b i 24
f
10a
2x
= 0.1
17a
1
= ; x = −3
8
1
5
2x = log10 (4); x ≈ 0.30
−x = loge (3); x ≈ −1.10
x3 = 125; x = 5
1
x−2 = 25; x =
5
2
−1
3
0
−1
0
10
−nlog4 (3)
1
6
1
log10 (2)
PA
iii
G
ii 9 = 3
10−1
11a
b
c
d
e
f
12a
b
c
d
13a
LI
N
f
N
e
O
d
E
b 25−0.5 = x; x =
c
b i log11 (18) ≈ 1.205
ii −log5 (8) ≈ −1.292
iii
620 1
2
7
4
5
6
23
1
2
0.001, 8
3
p+q
3q
2p + q
p−q
2q − 4p
3
pq
4
y = 100x
y = 22x × x−4
y = 23x
y = log10 (x) + 2
1
y=
x
y = x, x > 0
O
15a
O
FS
f 1.88
PR
ii
ii x < 0.5693
log7 (3) ≈ 0.2823
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
b
c
d
e
f
18a log2 (5), log2 (9)
b log5 (
5 − 2)
1
2
c 0, log9 (2) alternatively, x = 0, log3 (2)
d 2
1
, 16
4
f 1, 999
19a x ≈ 1.574
33 − 5
b x =
2
ln(5) + ln(2)
+ log(5); proof required
20a
ln(5)
e
b 1; proof required — check with your teacher
Exercise 11.4
1aFor y =
b
R+
and for y =
Asymptote is y = 0 for both graphs.
3x, range is
−3x, the range is
y = 5 – 4 × 3–x
0
(0, 1)
(–0.203, 0)
y = 3x
(0, 1)
y=5
y
R−.
x
0
range (−∞, 5)
6 a = −2, b = 2
y
7a i
y=0
x
(0, –1)
(1, 8)
y = –3x
2
y
(1, 1.5)
()
2x
y= –
3
y = (1.5)x
2
1, –
3
( )
(0, 1)
0
3 −x
=
3
3a
2
ii As the base increases, the graphs become
steeper (for x > 0).
y=0 x
and y =
(1.5) x
=
b i
2
x
E
G
(0, –91 )
y=0
x
E
0
LI
N
range R +
y
PA
y
(–2, 1)
4
(
x
)
N
17
0, —
16
O
0
y=1
y = 7–x
y = 5–x
(–1, 5)
y
(–1, 9)
(–1, 7)
(0, 1)
y=0 x
0
ii As the base increases, the graphs decrease more
steeply (for x < 0).
b i y = (0.8)x
y
(1, 1.25)
y = (0.8)–x
(–1, 0.8)
0
(1, 0.8)
y=0 x
(0, 1)
ii y = (1.25) x is the same as y = 0.8−x. Reflecting
these in the y-axis gives y = 0.8x.
range (1, ∞)
5a
y=0 x
0
ii y = 4−x; y = 6−x; y = 8−x
y = 4x–2 + 1
(2, 2)
(0, 1)
8a i y = 9–x
range (−2, ∞)
y = 3–(x + 2)
y
()
y = –2
b
1 x
y= –
8
(–1, 8)
PR
(0, −1)
()
x
(–1, 6)
1 x
y= –
4
(–1, 4)
y = 4x – 2
0
()
1
y= –
6
3 x
y
(0.5, 0)
y=0 x
O
y=
2 x
0
(1, 6)
y = 4x
(1, 4)
O
FS
b y =
(0, 1)
3−x
y = 8x
y = 6x
y
y = 1– × 101–2x
2
(0, 5)
0
( 1–2 , 1–2 )
y=0
x
range R +
Topic 11 Exponential functions 621
b (0, −9), (−2, −1)
y
9
x-intercept
Point
a
y=1
(0, 2)
none
(−1, 6)
b
y=1
(0, 0)
(0, 0)
(1, −3)
c
y = −2
(0, −1)
(log10 (2), 0)
(1, 8)
d
y = 6.25
(0, 5.25)
(−2, 0)
(−1, 4)
a (–2, –1)
y = –3x + 2
(0, –9)
y
(–1, 6)
y=0
x
0
c
0, 12 , (0.5, 1)
y = 5–x + 1
y
y = 4x – 0.5
O
FS
Asymptote y-intercept
(0, 2)
y=1
0
x
(0, 0.5)
x
d (0, 7), (1, 1)
(1, –3)
E
y
y = 10x – 2
y
0
y=0
x
11Each graph has asymptote at y = 0.
a (0, 3), (1, 6)
y = 6.25
E
y = 6.25 – (2.5)–x
y = 71 – x
(1, 1)
(0, 5.25)
y = 3 × 2x
(1, 6)
LI
N
d y = –2
x
PA
(log(2), 0)
G
(1, 8)
(0, –1)
y
(0, 7)
y = 1 – 4x
0
PR
0
O
y=1
(0, 0)
c y=0
x
0
y
b (0.5, 1)
(–1, 4)
(–2, 0)
x
N
0
O
10Each graph has asymptote at y = 0.
a
0, 14 , (2, 1)
0
b (0, 1),
y=0
x
4
,2
3
y
3x
–
y = 24
y
( )
1
0, –
4
0
622 (0, 3)
(0, 1)
y = 2x – 2
(2, 1)
y=0
x
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
0
( )
4, 2
–
3
y=0
x
14
y
y=0
0
x
(0, –3)
y = –3 × 2–3x
(– )
–
1 , –6
3
d (0, 1.5), (−2, 15)
y
Asymptote y-intercept x-intercept
a
y = −20
(0, −18)
(0.5, 0)
(−20, ∞)
b
y = −1
(0, 9)
(3.3, 0)
(−1, ∞)
c
y=3
(0, 1)
(−1, 0)
(−∞, 3)
d
y = −7
(0, 0)
(0, 0)
(−7, ∞)
e
y=8
(0, 7.2)
(0.7, 0)
(−∞, 8)
f
y = −4
(0, −4.2)
none
(−∞, −4)
a y
x
––
2
y = 1.5 × 10
y = 2 × 102x – 20
(0.5, 0)
x
0
(–2, 15)
(0, –18)
b (0, 9)
12aGraphs are identical with asymptote y = 0;
y = 5 × 21 – x – 1
y-intercept (0, 1); point (1, 9)
y
G
(1, 9)
(0, 1)
LI
N
b i y = 2x+1
y=3
()
2
y=3–2 –
3
(0, 1)
x
0
d y
N
y = 2x +1
(–1, 1)
O
(0, 2)
0
13a y = 2 ×
10x
y = 2(3.5)x + 1 – 7
(0, 0)
y=0
x
Asymptote y = 0; y-intercept (0, 2); point (−1, 1)
x
(–1, 0)
y
ii
x
y
c E
y=0
x
0
y = –1
PA
y = 9x or y = 32x
(3.3, 0)
0
E
y
PR
x
y = –20
O
(0, 1.5) y = 0
0
Range
O
FS
1
c (0, −3), −3, −6
x
(–1, –5)
y = –7
+3
b y = 4 ×
asymptote y = 0
c y = 6 − 2 × 31−x
d y = 6 − 6 × 3−x
32x;
Topic 11 Exponential functions 623
e y
3 log(y) = 2.5log(x) + 2; y = 100x2.5
y=8
(0, 7.2)
4 log(y) = 0.3x, y = 100.3x
y = 8 – 4 × 52x – 1
5a k =
1
5
b 10 years
6a 30 drosophilae
b 39 insects
(0, 7.0)
x
0
c 14 days
d
y
80
x
0
60
y = –2 × 10 3x –1 – 4
y = –4
1 , –6
–
3
20 (0, 30)
0
15a i −3
ii 3 − 3
2
y=3
0
(0, 3 –3 2)
x
b 5.6 %
(1, –3)
8a 7 °C
E
b Approximately
x–
–1
2 3
G
c
PA
(3, –9)
Asymptote y = 3; range (−∞, 3)
d x ≤ −0.17
16a One intersection for which x ∈ (−1, −0.5)
b Three
c One
d None
e One point of intersection (1, 6)
f Infinite points of intersection (t, 22t−1), t ∈ R
17 (−0.77, 0.59), (2, 4), (4, 16)
LI
N
N
Asymptote y-intercept
y = 33
O
y2
x-intercept
Point
(0, 31)
(1.17, 0)
(1.0185, 10)
(0, 11)
(0.17, 0)
(0.0185, 10)
T = 85 × 30.008t
(0, 85)
80
(31, 65)
(40, 60)
0
10
20
30
40
50
t
d Unrealistic for temperature to approach 0°
9a a = 70
b 77.6 °C
c 4 minutes
d
100 (0, 95)
80
60
(2, 78)
(4, 65)
T = 70 × 30.13t + 25
40
20
b k ≈ 0.004
0
c 5.3 kg
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
100
20
1a Proof required — check with your teacher
624 T
40
Exercise 11.5
b 16 weeks
hour
60
Horizontal translation is one unit to the left.
2a 42 emails per day on average
1
2
(10, 78)
E
y = 33
t
iii 7.45 years
(–1, 0)
y1
20
ii $2030.19
y
18
15
7a i Proof required — check with your teacher
iii Not possible
y=3–6×
10
e After 25 days
ii x = −1
c
5
PR
b i x = 3
(14, 60)
(5, 39)
40
( )
(0, –4.2)
N = 30 × 20.072t
O
FS
f O
N
100
T = 25
5
10
temperature approaches 25 °C
15
20
t
d i a = 0.10
10a k ≈ 0.054
ii p
b P0 ≈ 101.317
c 55.71 kilopascals; 76.80 kilopascals
(0.93, 80.20)
(0, 101.317)
100
(2.2, 76.8)
80
p = (200t + 16) × 2.7–t
P = 101.37 × 10–0.054h
(11.01, 37.07)
(4.8, 55.7)
(5.9, 48.7)
60
(8.8, 33.8)
40
t
20
Maximum turning points at
0
1
2
3
4
5
6
7
8
9
(0.93, 80.2), (11.01, 37.07)
10 11 h
Exercise 11.6
11a D0 × 1015k = 15 and D0 × 1018k = 75
b Proof required — check with your teacher
1a y = 10x
c 229 birds per square kilometre
y
O
1
5730
b 1540 years old
−
ii y = 2
x
0
4
b iBleach pH = 13, water pH = 7
ii 1 × 10−2 = 0.01, 1 × 10−6 = 0.000001
1.23
1.36
b log (y)
1.6
1.5
1.4
1.3
1.2
1.1
(53, 1.36)
y
y = 2–x
y=x
(0, 1)
PA
1.14
G
times less acidic; concentration of hydrogen ions
decreases by a factor of 10.
2 Inverse is y = −log2 (x)
E
iv An increase of one unit makes the solution 10
log(y)
x
(1, 0)
b log10 (1000) − log10 (100) = log10 (10) and 3 − 2 = 1
iii Lemon juice is 104 times more acidic
14a
y = log10 (x)
(0, 1)
x
PR
13a i y = 0.1
y=x
y = 10x
d Not lower than 39 birds per square kilometre
12a k =
O
FS
d
(1, 0) x
0
y = –log2(x)
E
(30, 1.23)
0
LI
N
(15, 1.14)
15
25
35
45
55
65
x
3a (−∞, 4)
1
b Rule is f −1 (x) = 3 log2 (4 − x); mapping is
f −1 : (−∞, 4) → R, f −1 (x) = 13log2 (4 − x)
c Proof required — check with your teacher
d Proof required — check with your teacher
c
y
N
e Approximately 50 years
y=4
f Model supports the claim.
O
15a i y ≈ 4 × 1.38x
ii y ≈ 4.3 + 6.2loge (x)
y = 3 (0, 3)
f
y=x
b Exponential model is the better model.
16a i 16
ii 51.5
b i 80.20
(0, 1)
0
x=4
f –1
(3, 0)
x
ii 55.61 seconds
iii 7.08; 0.10
c 10 minutes
Topic 11 Exponential functions 625
4a y = 2
y
b
y
9a
x−4
2
y = 5x
y = 3x
(1, 5)
f
y=x
f
y=x
–1
(1, 3)
x
0
x = 3y
(3, 1)
(1, 0)
(0, 1)
x
b 0.001
1
x
6 × 2
3
7aAsymptote x = 1; domain (1, ∞); x-intercept (2, 0);
x=1
O
FS
0
c Twice
5a 2x
b y = log3 (x), y = log5 (x)
y = log10(x – 1)
c The larger the base, the more slowly the graph
(2, 0)
d
x
increases for x > 1.
y
b Asymptote x = 0; domain R + ; x-intercept (5, 0);
y x=0
E
PA
c i b = 1
ii Domain R; range (−1, ∞), f −1 (x) = 2−x − 1
y
iii
E
y = 5 x +1
(0, 5)
x
y=x
y = –1
(–1, 1)
Inverse
0
(5, 0)
O
(1, –1)
of 1 unit upwards
y
y = log5 (x) − 1
10a y = −3log2 (x)
y = 1 – log4 (x)
y = log4 (x)
(4, 0)
(1, 0)
x=1
f
Asymptote y = −1, contains the origin
8 Reflection in the x-axis followed by a vertical translation
(2, 0.5)
x
(2, 0)
e y = log3 (x − 1)
N
LI
N
(0, 0)
x = 3y +1
0
G
y = log5 (x) – 1
–1(x)
Inverse
y=1
x
0
x
y = – log4 (x)
Asymptote x = 0, domain R + , x-intercept (4, 0)
626 y=x
(0, 2)
(5, 0)
y=f
y = 3x+1
PR
O
0
0
(5, 1)
x = 5y
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
y=2
y
x
––
3
(–3, 2)
Inverse
y=x
(0, 1)
0 (1, 0)
(2, –3)
x
x
12a x = 2
b y = log8 (2x + 2)
x = –1
1
0, –
3
( ) ( )
1, 0
––
2
b y = 4.5
1 × x
y= –
2 8 –1
y
Inverse
y=x
Inverse
0
( )
( )
(0, 4.5)
x
(2, 2)
(4.5, 0)
1
0, –
2
y=2
0
f x = 2
y=x
(1, 0)
( )
1
–
,0
2
Inverse
x = 2 – 4x
13aRange (−1, ∞)
x
x=2
b 0
c (0, 0)
d y = log3 (x − 3) − 1
PR
(0, 6)
y = g–1 (x)
Inverse
(6, 0)
(4, –1)
1
e y = −2log10 (−x)
Inverse
(0, 1)
(–1, 0)
y
f
x = –1
(0, 0)
(4, –3)
x
LI
N
(0, –1)
y
x = –4
N
O
y=x
0, 1–2
0
y=x
(1, 1)
Inverse
( )
x
1– , 0
2
(0, –2)
x
0 (2, 0)
y = 22x + 1 – 4
( )
(–2, 0)
y
21 – x
x
y = g–1 (x)
f y = 1 − log2 (x)
(0, 2)
x
y = –1
(1, 0)
0
y=x
(0, 0)
E
y
x
PA
0
y
E
y=3
−1
(–3, 4)
y=x
G
(–1, 4)
e
x=3
x
3
−
d g−1 : R → R, g−1 (x) = 5
y
y = 3x + 1 + 3
x
2
e y = 2 − log1.5
(0, 1)
x
O
FS
( )
0
y
O
y=
y = 2 × (1.5)2 – x
1
0, – –
2
c y = log4 (2 − x)
y=x
1, 0
–
3
y = –1
y
c and d Inverse
y = –4
11aDomain (−∞, 8); range R
b 6
c (6, 0)
8−x
1
log3
; mapping
2
2
8−x
1
is f −1 : (−∞, 8) → R, f −1 (x) = log3
.
2
2
d Rule is f −1 (x) =
14a i 8
ii 6
b i x
15a y =
ii
x3
iii 0.5
iv 5
iii 3x
iv x
−4
1
1
b y = log2 (x + 4) −
2
2
22x+1
Topic 11 Exponential functions 627
c Asymptote: x = −4; intercepts: (−2, 0), (0, 0.5)
y
f d Two points of intersection
x = –4
e k = 14
y = –log2 (x + 4)
(–3, 0)
f Proof required — check with your teacher
16Range is R for each graph.
asymptote equation
axes intercepts
a
R+
x=0
(25, 0)
b
(2, ∞)
x=2
(3, 0)
c
R+
x=0
(0.1, 0)
d
(−1, ∞)
x = −1
e
(−∞, 4)
x=4
f
(−4, ∞)
x = −4
(0, 0)
(3, 0), (0, log3 (4))
(−3, 0), (0, −2)
reflection in y-axis; horizontal shift 2 units to right
b Vertical translation of 1 unit up
1
c Dilation of factor
from the x-axis
log10 (2)
Proof required — check with your teacher
x
2
b i y = −4 log3 (x) + 4
E
y = log5 (x – 2)
ii (0, 3)
x
c i y = −log2 (x + 2) − 1
G
0
x=0
PA
y
y = log10 (x) + 1
(1, 1)
x
(0.1, 0)
ii y = 2−(x+1) −2
y
Inverse
(–2, 0)
LI
N
E
0
Proof required — check with your teacher
18a y = 14 log7
y
x=2
c n
O
x
(2, b log2 (3))
y = log5 (x) –2
(25, 0)
b t = b log2 (n + 1)
n = –1
(0, 0)
x=0
0
t
d
PR
a y
17aDilation factor 2 from the x-axis; reflection in x-axis;
O
FS
domain
x
(0, –2)
(0, –1.5)
y = –2
y
d x
0
y = log10(x + 1)
x
(0, 0)
e y = log3 (4 – x)
9
4
1
2
iv Sketch here graphs using a CAS technology
19 (0.4712, 4.632), (2, 2), (4.632, 0.4712)
iii f : (−2, 0), (0, 2); g : (10, 0), 0,
y
x=4
(0, log3 (4)
(3, 0)
0
628 9
4
d i df = − , ∞ ; dg = (−∞, 20)
ii x = − , x = 20
O
N
x = –1
x
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
20a i Graphs are not the same.
y
c
x=0
4
ii x =
3
b
y
x=0
0
(–1, 0)
0
(1, 0)
x
x
(1, 0)
domain R \ 0 ; range R; many-to-one correspondence
d The law holds for m > 0 and it does hold since
log3 (x2) = 2log3 (x), x > 0.
domain R + ; range R; one-to-one correspondence
O
N
LI
N
E
PA
G
E
PR
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Topic 11 Exponential functions 629