264
57.
5
61.
6
65.
x
58.
x5
62.
8
d4 x 3 n
66.
1
3
x
1
4
63.
x3
10
d3 x 5 n
69.
4
x 6 x 10
70.
6
73.
3
9 6
x x
74.
2
77.
c
78.
x19 4
c 4m
x
1
x9 3
m
x3
1
59.
67.
x 8x 4
10 - 6
x x
1
x3
1
3
x4
1
(2x)– 3
3
75.
^27x6h3
x 9 y 12
x2
64.
5
x3
72.
1
(3x)– 4
4
x 12 y 8
1
1
^81x8h 4
76.
^- xh3 ^- xh4
x
4
68.
71.
79.
60.
^- xh6 ^- xh4
80.
5
x9
8.4 Logarithms
Concept of Logarithm and its Relation to Exponents
In the previous section, power of a variable (or term or number) was introduced using exponents and bases.
x
Power, a = a : a : a : ... a
x factor of a
Exponent
ax = y
Power
Number
Base
Read as: base ‘a ’ raised to the ‘x ’ is
‘y ’ or ‘a ’ to the power of ‘x ’ is ‘y ’.
In exponential form
Base
Exponent
ax = y
a
x
103 = 1,000
10
3
52 = 25
5
2
In this section you will learn how to determine the exponent, x, to which the base, a, must be raised
to obtain y; i.e., to determine the exponent of the base, a, to produce y.
Logarithm is defined as the exponent, x, to which a base, a, must be raised to obtain y.
Number
Loga y = x
Logarithm
(exponent)
Base
Read as: logarithm of ‘y ’ to the base ‘a ’
is ‘x ’ or log ‘y ’ to the base ‘a ’ is ‘x ’.
In logarithmic form
Base
Exponent
In
exponential
form
loga y = x
a
x
ax = y
log10 1,000 = 3
10
3
103 = 1,000
log5 25 = 2
5
2
52 = 25
Exponent
x
a =y
Exponential
notation
Chapter 8 | Basic Algebra
loga y = x
Base
Logarithmic
notation
x
a
x
loga y
y
Exponents and logarithms are inverse functions.
265
For example,
• 103 = 1,000
3
log10 1,000 = 3
103
log10 1,000
2
• 5 = 25
2
1,000
log5 25 = 2
2
5
log5 25
25
As seen above, a logarithm is the exponent to which the base is raised to get a number.
Any positive number can be used as the base for logarithms.
For example, 100 is the same as 102. Here, the base is 10 and the exponent is 2. Therefore, the logarithm
of 100 to the base 10 is 2.
log10 100 = 2
102 = 100
1 44
4 2 44
43
1 44 2 44 3
is the same as Logarithmic form
Exponential form
3
Similarly, 125 is the same as 5 . Here, the base is 5 and the exponent is 3; therefore, the logarithm of
125 to the base 5 is 3.
53 = 125
14 24 3
Example 8.4-a
Exponential form
is the same as
log5 125 = 3
1 44
2 44 3
Logarithmic form
Converting Exponential Form to Logarithmic Form
Convert the following to logarithmic form:
(i)
2
6 = 36
1
32
(iv) 2–5 =
Solution
x
Using a = y
(i)
2
6 = 36
log6 36 = 2
–5
(iv) 2 =
log2
Example 8.4-b
(ii) 26 = 64
1
32
1
= –5
32
(v)
(iii) 73 = 343
ab = c
loga y = x,
(ii) 26 = 64
(iii) 73 = 343
log2 64 = 6
log7 343 = 3
(v)
ab = c
loga c = b
Converting Logarithmic Form to Exponential Form
Convert the following to exponential form:
(i)
log3 81 = 4
(iv) log2 d
Solution
(ii) log2 128 = 7
1
n = –4
16
Using loga y = x
(v) loga b = c
x
a = y,
(ii) log2 128 = 7
(iii) log3 2,187 = 7
34 = 81
27 = 128
37 = 2,187
(iv) log2 d
(v) loga b = c
(i)
log3 81 = 4
(iii) log3 2,187 = 7
1
n = –4
16
1
2–4 =
16
ac = b
8.4 Logarithms
266
­­­Common Logarithms (log)
Common logarithms are always to the base 10. If no base is shown in the common logarithmic
expression, it is assumed to be of base 10 and is referred to by the symbol 'log'.
Common Logarithmic Form
Exponential Form
log101,000 = log 1,000 = 3
103 = 1,000
log1010 = log 10 = 1
101 = 10
log10100 = log 100 = 2
102 = 100
log101 = log 1 = 0
100 = 1
x
log10 y = log y = x10 = y
Example 8.4-c
Finding Common Logarithm of Numbers
Calculate the following, rounding to 4 decimal places.
(i)
Solution
log 10,000
Using
(i)
(ii) log 40
(iii) log 6.5
(iv) log 0.25
button on the calculator,
log 10,000 = 4
(ii) log 40 = 1.602059... (iii) log 6.5 = 0.812913... (iv) log 0.25 = –0.602059...
= 1.6021
= 0.8129
= –0.6021
­­­Natural Logarithms (ln)
Natural logarithms are always to the base ‘ e ’ where the constant e = 2.718282...
'e' is a special number in mathematics (similar to π, which is equal to 3.141592...) and is found by
1 n
`1 + n j , where ‘ n ’ is a large number.
100, 000
1
= 2.718282...
Assume n = 100,000; therefore, e = c1 + 100, 000 m
In business and financial calculators, the natural logarithm key, 'ln', is the only logarithmic key
available. The common logarithm key, log, is not available. Natural logarithm is referred to by the
symbol 'ln' (pronounced "lawn"), which has the base 'e'.
If the base of a logarithmic expression is ‘ e ’, then it is simply expressed by the symbol 'ln'.
We know x 0 = 1
Thus, 100 = 1
and e0 = 1
Exponential Form
loge1 = ln 1 = 0 e 0 = 1
loge e = ln e = 1 e 1 = 2.718282…
loge 1.005 = ln 1.005 = 0.00498754... e 0.00498754 = 1.005
loge 10 = ln 10 = 2.302585... Therefore,
log101 = 0
lne1 = 0
Natural Logarithmic Form log 1 = 0
ln 1 = 0
Example 8.4-d
e 2.302585 = 10
The rules of logarithms are used to evaluate the exponent ‘n’ in business and finance mathematics formulas.
Finding Natural Logarithm of Numbers
Calculate the following, rounding to 4 decimal places.
(i)
Chapter 8 | Basic Algebra
ln 1,000
(ii) ln 50
(iii) ln 1.05
(iv) ln 0.50
267
Solution
Using the
(i)
button on the calculator,
ln 1,000 = 6.907755... (ii) ln 50 = 3.912023...
= 6.9078
= 3.9120
(iii) ln 1.05 = 0.048790... (iv) ln 0.50 = –0.693147...
= 0.0488
= –0.6931
Rules of Logarithms
Table 8.4
Rules of Logarithms
Rule
Description
Rule in Common Logarithmic Form
Product Rule
Logarithm of a product equals
the sum of the logarithms of the
factors.
loga (AB) = loga A + loga B
Quotient Rule
Logarithm of a quotient equals
the difference between the
logarithms of the numerator
(dividend) and the logarithm of
the denominator (divisor).
loga d A n = loga A – loga B
B
3.
Power Rule
Logarithm of a number raised to
a power equals the product of the
power and the logarithm of the
number.
n
loga (A) = n loga . A
4.
Logarithm of 1 Rule
Logarithm of 1 is zero.
loga 1 = 0
5.
ogarithm of the
L
Base Rule
Logarithm of the base is one.
loga a = 1
6.
ogarithm of a
L
Base Raised to a
Power Rule
Logarithm of the base raised to a
power is equal to the power.
loga (a)n = n
7.
Logarithm of
Roots (radicals)
Rule
Convert the roots (radicals) to
exponents and then apply the
Power Rule.
loga (n A ) = loga (A)n = 1 . loga A
n
1.
2.
1
Note:
loga (M + N) ≠ loga M + loga N
loga (M – N) ≠ logaM – loga N
(loga M)(loga N) ≠ loga M + loga N
loga M
≠ loga M – loga N
loga N
1. Rules of logarithms can be used to combine two or more logarithmic expressions into a single
logarithmic expression.
2. Common logarithms (log) and natural logarithms (ln) follow the same rules.
3. To change the base of a logarithm, use the following formula:
logc b
loga b =
, (i.e., changing from base 'a' to base 'c'.)
logc a
Example: loge A =
log10 A
log A
(i.e., ln A =
)
log e
log10 e
log10 A =
loge A
ln A
(i.e., log A =
).
loge 10
ln 10
8.4 Logarithms
268
Example 8.4-e
­­Evaluating Logarithms Using Rules of Logarithms
Evaluate the following common logarithms.
Solution
(ii) log d
(i)
log (100 × 100)
(i)
log (100 × 100)
= log 100 + log 100
=2+2
=4
75
n
50
(iii) log 1255
Using Product Rule,
(iv) log 109
log (100 × 100)
Using exponents,
= log (100) Using Power Rule,
2
or
= 2 . log (100)
= 2 . (2)
=4
75
(ii) log d 50 n
Using Quotient Rule,
= log 75 – log 50
= 1.875061... – 1.698970...
= 0.176091... = 0.1761
5
Using Power Rule,
(iii) log (125)
= 5 . log (125)
= 5 . (2.09690...)
= 10.484550... = 10.4846
(iv) log (10)9
Example 8.4-f
= 9 . log (10)
= 9 . (1)
=9
Using Power Rule,
­­Evaluating Logarithms Using Rules of Logarithms
Evaluate the following using natural logarithms.
Solution
(i)
ln (275 × 75)
4,750
(ii) ln d 3,275 n
(i)
ln (275 × 75) Using Product Rule,
= ln 275 + ln 75
= 5.616771... + 4.317488...
= 9.934259... = 9.9343
6
(iii) ln (4.25) Using Power Rule,
(iii) ln (4.25)6
(ii) ln d
4,750
n
3,275 Using Quotient Rule,
= ln 4,750 – ln 3,275
= 8.465899... – 8.094073...
= 0.371826... = 0.3718
(iv) ln (e)4 Using Power Rule,
= 6 . ln (4.25)
= 4 . ln (e)
= 6 . (1.446918...)
= 4 . (1) = 4
= 8.681513... = 8.6815
or
ln (e)4 4
= loge (e)
Chapter 8 | Basic Algebra
(iv) ln (e)4
=4
Using Logarithm of the Base Rule,
269
Example 8.4-g
­­Writing Single Logarithms
Write each of the following as a single logarithm and evaluate.
Solution
(i)
2 log 3 + log 5
(ii) 3 log 4 – log 8
(i)
2 log 3 + log 5
(ii) 3 log 4 – log 8
= log 3 +log 5
2
= log (32 × 5)
= log 45
= 1.653212... = 1.6532
5
= log 2 + log 4 – log 8
= log 32 + log 4 – log 8
= log (32 × 4 )
8
= log 16
= 1.204119... = 1.2041
Example 8.4-h
= log (4)3 – log 8
= log c 4 m
8
= log 8
= 0.903089... = 0.9031
3
(iii) 5 log 2 + log 4 – log 8
(iv) log 3 216
(iii) 5 log 2 + log 4 – log 8
(iv) log 3 216
1
= log (216) 3
= 1 log 216
3
= 1 (2.334453...)
3
= 0.778151... = 0.7782
­­Solving Equations
Solve for ‘ n ’ in the following equations:
(i) 1,024 = 2n
(ii) 3,749 = 1,217(1.005)n
Solution
Using natural logarithms:
1,024 = 2n
(i)
ln 1,024 = ln 2n
Using common logarithms:
n
1,024 = 2 Taking ln on both sides,
log 1,024 = log 2 Using Power Rule,
Using Power Rule,
or log 1,024 = n $ log 2
log 1,024
n=
log 2
= 3.010299...
0.301029...
ln 1,024 = n ln 2
Isolating n,
ln 1, 024
n = ln 2
= 6.931471...
0.693147...
= 10
(ii)
ln c
3,749 = 1,217(1.005)n Dividing both sides by 1,217,
3, 749 = (1.005)n
1, 217
3, 749
ln c
= ln (1.005)n
1, 217 m
Taking log on both sides,
n
= 10
3,749 = 1,217(1.005)n Dividing both sides by 1,217,
Taking ln on both sides,
n
3,749 = (1.005)
1,217
Using Power Rule,
log d
3, 749
= n ln 1.005
Isolating n,
1, 217 m
ln c 3, 749 m
1, 217
n=
ln 1.005
1.125100f
=
0.004987f
or
Isolating n,
3,749
n = n . log (1.005)
1,217
3,749
log d 1,217 n
n=
log 1.005
=
Taking log on both sides,
Isolating n,
0.488624...
0.002166...
= 225.582147... = 225.5821
= 225.582147... = 225.5821
8.4 Logarithms
270
8.4 Exercises
Answers to odd-numbered problems are available at the end of the textbook.
Express the following in logarithmic form:
1.
a. 105 = 100,000
b. 45 = 1,024
2.
a. 104 = 10,000
b. 44 = 256
3.
a. 26 = 64
b. 65 = 7,776
4.
a. 23 = 8
b. 64 = 1,296
5.
a. 32 = 9
b. 94 = 6,561
6.
a. 33 = 27
b. 82 = 64
Express the following in exponential form:
7.
a. log10100 = 2
b. log464 = 3
8.
a. log101,000 = 3
b. log44,096 = 6
9.
a. log232 = 5
b. log5625 = 4
10.
a. log24 = 2
b. log5125 = 3
11.
a. log3729 = 6
b. log6216 = 3
12.
a. log3243 = 5
b. log61,296 = 4
Calculate the following (rounding to 4 decimal places):
13.
a. log 225
b. log 1.54
14.
a. log 27
b. log 2.5
15.
a. log 35
b. log 0.25
16.
a. log 155
b. log 0.75
17.
a. ln 10.05
b. ln 1.005
18.
a. ln 0.675
b. ln 750
19.
a. ln 0.165
b. ln 1.02
20.
a. ln 12.51
b. ln 72
Solve for 'n' (rounding to 2 decimal places):
n
21.
250 = (30)
25.
7,500 = (45)n + 500
22.
320 = (15)n
23.
(1.05)n = 1.31
24.
2.5 = (1.05)n
26.
8,000 = (35)n + 1,500
27.
10,000 = 2,000(1.2)n
28.
15,000 = 5,000 (1.04)n
Express the following as a sum or difference of two or more natural logarithms:
29.
ln ` 3 j
7
30.
ln ` 40 j
13
31.
ln (4 × 9)
32.
ln (7 × 8)
33.
ln ` AB j
C
34.
ln ` x j
ab
35.
ln ` X j
YZ
36.
ln `
37.
ln c 3x m
2yz
38.
ln ` 5x j
2ab
39.
ln c
xy
m
z
40.
ln c
44.
loga (x 2y)
48.
loga (yx)– 2
52.
loga d
xy
j
c
x
m
yz
If loga x = M and loga y = N, express the value of each of the following in terms of M and N.
41.
45.
49.
loga d x m
y
2
x
loga d m
y
loga (5 x 4 )
42.
46.
50.
loga d y m
x
loga d x2 m
y
loga d 3
1
m
x2 y 2
43.
loga (xy 2)
47.
loga (xy)– 2
51.
loga d3
1
1
m
xy
1
1
m
x y3
Express the following as a single natural logarithm:
53.
ln 8 + ln 5
54.
ln 25 + ln 4
55.
ln 15 − ln 3
56.
ln 60 − ln 15
57.
2 ln 5 + 3 ln 3
58.
2 ln 8 + 3 ln 3
59.
5 ln2 − 2 ln 3
60.
4 ln 5 − 3 ln 2
61.
2 ln 5
62.
5 ln 2
65.
2 lnd x m
2 ln
y
69.
3 ln a + 2 ln b − 5 ln c 70.
66.
5 lnd a m
5 ln
b
63.
3 ln 6
64.
6 ln 3
67.
4 ln (a × b)
68.
3 ln (xy)
4 ln x − 2 ln y + 3 ln z
71.
3 ln 2 + 4 ln 3 − 2 ln 4
72.
2 ln 3 + 3 ln 2 − 4 ln 2
Express the following in the form K log10 M, simplify, and then evaluate to 2 decimal places, wherever applicable.
73.
log10 1,000
74.
log10 4 10,000
75.
log10 36
76.
log10 625
77.
log10 4 81
78.
log10 3 216
79.
log10 3 9 2
80.
log10 5 323
Chapter 8 | Basic Algebra
271
Solve for ‘n’ (rounding to 2 decimal places):
= ln c 4, 285 m
81. nn =
4, 000
7, 200
m
4, 725
85. nn==
ln ^1.01h
ln c
82.
= ln c
nn =
6, 750
m
3, 200
83.
= ln c
nn =
86.
5, 120
m
2, 250
=
nn =
ln ^1.005h
87.
nn==
ln c
3, 645
m
2, 175
ln ^2.5h
ln ^1.03h
84. nn== ln c
=
88. nn =
75, 000
m
2, 200
ln ^3 h
ln ^1.02h
8.5 Rearranging Equations and Formulas
Introduction
Equations are mathematical statements formed by placing an equal (=) sign between two expressions
to indicate that the expression on the left side is equal to the expression on the right side of the equal sign.
5x + 3 = y – x
For example, Formulas are similar to equations. In formulas, the relationship among many variables is written as a
rule for performing calculations. Formulas are written so that a single variable, known as the subject
of the formula, is on the left side of the equation, and everything else is on the right side.
I = Prt
For example,
Isolating Variables
To isolate a particular variable in an equation or a formula, rearrange the terms and simplify, so that
the required variable is on the left side of the equation and all the other variables and numbers are
on the right side of the equation. Rearrangement can be performed by using the rules that you have
learned in the previous sections of this chapter and the following guidelines:
■■ Add or subtract the same quantity to or from both sides.
■■ Multiply or divide both sides by the same quantity.
■■ Take powers or roots on both sides.
■■ Expand the brackets and collect the like terms.
■■ Remove the fractions by multiplying both sides by the denominator or LCM.
For example, consider the formula for simple interest: I = Prt.
To solve for any of the variables, 'I', 'P', 'r', or 't' in this simple interest formula, we can rearrange the
variables as shown below:
I = Prt is the same as Prt = I
Solving for 'P':
Solving for 'r':
Solving for 't':
Prt
I
rt = rt I
P = rt
Prt
= I Pt
Pt
r= I
Pt
Prt
I
=
Pr
Pr
Dividing both sides by 'rt',
Dividing both sides by 'Pt',
Dividing both sides by 'Pr',
t = I
Pr
8.5 Rearranging Equations and Formulas