Higher Unit 3
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Higher
Outcome 3
Exponential & Log Graphs
Special “e” and Links between Log and Exp
Rules for Logs
Solving Exponential Equations
Experimental & Theory
Harder Exponential & Log Graphs
Exam Type Questions
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The Exponential &
Logarithmic Functions
Outcome 3
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Higher
Logarithmic Graph
Exponential Graph
y
y
(0,1)
x
(1,0)
x
A Special Exponential Function
– the “Number” e
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Higher
Outcome 3
The letter e represents the value 2.718…..
(a never ending decimal).
This number occurs often in nature
f(x) = 2.718..x = ex
is called the exponential function to the base e.
Linking the Exponential
and the Logarithmic Function
Outcome 3
Higher
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y
f ( x)  2 x
f 1 ( x) 
(0,1)
log 2 x
(1, 0)
x
In Unit 1 we found
that the exponential
function has an
inverse function,
called the
logarithmic function.
log a 1  0
log a a  1
y  a x  x  log a y
The log function is the inverse of the exponential
function, so it ‘undoes’ the exponential function:
Linking the Exponential
and the Logarithmic Function
Outcome 3
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Higher
f(x) = 2x
ask yourself:
1
2
21 = 2
so
2
4
22 = 4
so
log24 = 2
3
8
23 = 8
so
log28 = 3
4
16
24 = 16
so
f(x) = log2x
log22 = 1
log216 = 4
“2 to what power gives 2?”
“2 to what power gives 4?”
“2 to what power gives 8?”
“2 to what power gives 16?”
Linking the Exponential
and the Logarithmic Function
Outcome 3
Higher
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f(x) = 2x
ask yourself:
1
2
21 = 2
so
2
4
22 = 4
so
3
8
23 = 8
so
4
16
24 = 16
so
log22 = 1
log24 = 2
log28 = 3
log216 = 4
“2 to what power gives 2?”
“2 to what power gives 4?”
“2 to what power gives 8?”
“2 to what power gives 16?”
f(x) = log2x
Examples
(a)
log381 = 4
1
2
(b)
log42 =
(c)
 1 
log3   = -3
 27 
“ 3 to what power gives 81 ?”
“ 4 to what power gives 2 ?”
“ 3 to what power
 1 
gives  27 
 
?”
Rules of Logarithms
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Higher
Outcome 3
Three rules to learn in this section
log a xy  log a x  log a y
x
log a    log a x  log a y
 y
log a x  p log a x
p
Rules of Logarithms
Higher
Outcome 3
Examples
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Simplify:
a)
log102 + log10500
b) log363 – log37
 63 
 log 3  
 7 
 log10 (2  500)
Since
log a a  n log a a
n
Since
log a a  1
 log10 1000
Since
log a a  n log a a
n
 log10 (10)3
 log3 9
 log 3 (3)
Since
2
 3  log10 10
 2  log3 3
3
2
log a a  1
Since
n
n log a b  logRules
b
a
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Higher
of Logarithms
Outcome 3
Example
1
1
log 2 16  log 2 8
2
3
Simplify
 log 2 (16)
1
2
 log 2 (8)
1
3
x
log a x  log a y  log a  
 y
Since
 log 2 4  log 2 2
4
 log 2  
2
 log 2 2
log a a  1
1
Using your Calculator
Outcome 3
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Higher
You have 2 logarithm buttons on your calculator:
log
ln
which stands for log10 and its inverse
which stands for loge
and its inverse
Try finding log10100 on your calculator
10x
log
ex
ln
2
Logarithms & Exponentials
Outcome 1
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Higher
We have now reached a stage where trial and error is
no longer required!
Solve ex = 14
(to 2 dp)
Solve ln(x) = 3.5 (to 3 dp)
ln(ex) = ln(14)
elnx = e3.5
x = ln(14)
x = e3.5
x = 2.64
Check e2.64 = 14.013
13 July 2017
x = 33.115
Check ln33.115 = 3.499
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Logarithms & Exponentials
Outcome 1
Higher
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Solve
3x = 52
( to 5 dp )
ln3x = ln(52)
xln3 = ln(52)
(Rule 3)
x = ln(52)  ln(3)
x = 3.59658
Check:
13 July 2017
33.59658 = 52.0001….
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Solving Exponential Equations
Higher
Example
Outcome 3
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Solve
Since
log a bn  n log a b
5  11
x
51 = 5 and 52 = 25
so we can see that x lies between 1 and 2
Taking logs of both sides and applying the rules
log10 5x  log10 11
x log10 5  log10 11
log10 11
x
 1.489
log10 5
Solving Exponential Equations
Higher
Outcome 3
Example
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For the formula P(t) = 50e-2t:
a)
Evaluate P(0)
b)
For what value of t is P(t) = ½P(0)?
(a)
P(0)  50e20  50
Remember
a0 always
equals 1
Solving Exponential Equations ln = loge e
Higher
Example
Outcome 3
logee = 1
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For the formula P(t) = 50e-2t:
b)
For what value of t is P(t) = ½P(0)?
1
1
P(0)   50  25
2
2
0.693  2t ln  e 
25  50e2t
0.693  2t 1
1
 e 2 t
2
0.693
t
2
1
ln    ln  e 2t 
2
t  0.346
Solving Exponential Equations
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Higher
Outcome 3
Example
The formula A = A0e-kt gives the amount of a
radioactive substance after time t minutes.
After 4 minutes 50g is reduced to 45g.
(a) Find the value of k to two significant figures.
(b) How long does it take for the substance to
reduce to half it original weight?
(a)
t 4
A(0)  50
A(4)  45
Solving Exponential Equations
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Higher
(a)
t 4
loga  xy   loga x  loga y
Outcome 3
Example
A(0)  50
 45  50e
A(4)  45
4 k
 ln(45)  ln  50e
4 k

 ln(45)  ln 50  ln e
4 k
 ln 45  ln 50  ln e4 k
Solving Exponential
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Higher
Example
x
log a x  log a y  log a  
Equations
 y
Outcome 3
ln 45  ln 50  ln e
 45 
ln    4k ln e
 50 
0.1054  4k
0.1054
k
4
k  0.0263
4 k
ln = loge e
logee = 1
Solving Exponential Equations
Higher
Example
Outcome 3
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(b) How long does it take for the substance to
reduce to half it original weight?
1
 A(0)  A(0)e 0.0263t
2
1
 ln    ln  e 0.0263t 
2
0.693  0.0263t ln  e 
0.693  0.0263t
1
  e 0.0263t
2
ln = loge e
logee = 1
 t  26.35 minutes
Experiment and Theory
Outcome 3
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Higher
When conducting an
experiment scientists may
analyse the data to find if a
formula connecting the
variables exists.
Data from an experiment
may result in a graph of the
form shown in the diagram,
indicating exponential
growth. A graph such as this
implies a formula of the type
y = kxn
y
x
Experiment and Theory
Outcome 3
Higher
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We can find this formula by using logarithms:
y  kx
If
log y
n
Then
log y  log kx
So
log y  log x n  log k
n
(0,log k)
log y  n log x  log k
Compare this to
So
Y  mX  c
log y  n log x  log k
log x
log y  Y
nm
log x  X
log k  c
Is the equation
of a straight line
Experiment and Theory
Outcome 3
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Higher
y  kx
From
log y
(0,log k)
log x
We see by taking logs that
we can reduce this problem
to a straight line problem
where:
log y  n log x  log k
And
n
Y = m X
+
c
Experiment and Theory
Outcome 3
Higher
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NB: straight line with
gradient 5 and intercept 0.69
ln(y)
Using
0.69
m=5
Y = mX + c
ln(y) = 5ln(x) + 0.69
ln(y) = 5ln(x) + ln(e0.69)
ln(y) = 5ln(x) + ln(2)
ln(x)
Express y in terms of x.
ln(y) = ln(x5) + ln(2)
ln(y) = ln(2x5)
y = 2x5
Experiment and Theory
Outcome 1
Higher
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log10y
0.3
Using
m =
-0.3/
1
= -0.3
Y = mX + c
Taking logs
log10y = -0.3log10x + 0.3
log10y = -0.3log10x + log10100.3
1
log10x
log10y = -0.3log10x + log102
log10y = log10x-0.3 + log102
Find the formula
connecting x and y.
log10y = log102x-0.3
straight line with intercept 0.3
y = 2x-0.3
Experimental Data
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Higher
Outcome 1
When scientists & engineers try to find relationships
between variables in experimental data the figures are
often very large or very small and drawing meaningful
graphs can be difficult. The graphs often take
exponential form so this adds to the difficulty.
By plotting log values instead we often convert from
The variables Q and T are known
to be related by a formula in the form
Outcome 3
Higher
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T = kQn
The following data is obtained from
experimenting
Q
5
T
300
10
15
20
25
5000 25300 80000 195300
Plotting a meaningful graph is too difficult so
taking log values instead we get ….
log10Q
0.7
1
1.2
1.3
1.4
log10T
2.5
3.7
4.4
4.9
5.3
m =
5.3 - 2.5
1.4 - 0.7
= 4
Point on line (a,b) = (1,3.7)
log10T
log10Q
Experiment and Theory
Higher
Outcome 3
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Since the graph does not cut the y-axis use
Y – b = m(X – a)
where X = log10Q and Y = log10T ,
log10T – 3.7 = 4(log10Q – 1)
log10T – 3.7 = 4log10Q – 4
log10T = 4log10Q – 0.3
log10T = log10Q4 – log10100.3
log10T = log10Q4 – log102
log10T = log10(Q4/2)
T = 1/2Q4
Experiment and Theory
Higher
Outcome 3
Example
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The following data was collected during an experiment:
X
50.1
194.9
501.2
707.9
y
20.9
46.8
83.2
102.3
a) Show that y and x are related by the formula y = kxn
.
b) Find the values of k and n and state the formula that
connects x and y.
X
50.1
194.9
501.2
707.9
y
20.9
46.8
83.2
102.3
Outcome 3
Higher
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a)
Taking logs of x and y and plotting points we get:
log10 Y
3
Since we get a
straight line the
formula connecting
X and Y is of the form
2.5
2
1.5
1
0.5
0.5
1
1.5
2
2.5
3
log10 X
Y  mX  c
Experiment and Theory
Outcome 3
Higher
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b)
Since the points lie on a straight line,
formula is of the form:
y  kx
Graph has equation
Compare this to
n
log y  n log x  log k
Y  mX  c
Selecting 2 points on the graph and substituting them
into the straight line equation we get:
Experiment and Theory
Higher
Outcome 3 ( any will do ! )
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The two points picked are  2.84, 2.00 and 1.69,1.32
2.00  2.84m  c (A)
1.32  1.69m  c (B)
Sim. Equations
Solving we get
0.68  1.15m
0.68
m
1.15
m  0.6
Sub in B to find value of c
1.32  1.69  0.6  c
c  0.3
Experiment and Theory
Outcome 3
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Higher
Y  0.6 X  0.3
So we have
Compare this to
log y  n log x  log k
so
n
0.6
and
log k  0.3
Experiment and Theory
Outcome 3
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Higher
solving
log k  0.3
k  10
You can
always
check this
on your
graphics
calculator
0.3
k  1.99
so
y  kx  y  1.99 x
n
0.6
Transformations of
Log & Exp Graphs
Higher
Outcome 3
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In this section we use the rules from Unit 1 Outcome 2
Here is the graph of y = log10x.
Make sketches of
y = log101000x and
y = log10(1/x) .
Graph Sketching
Outcome 3
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Higher
log101000x = log101000 + log10x = log10103 + log10x
= 3 + log10x
If f(x) = log10x then this is f(x) + 3
(1,3)
(10,4)
y = log101000x
(10,1)
(1,0)
y = log10x
Graph Sketching
Outcome 3
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Higher
log10(1/x) = log10x-1 = -log10x
If f(x) = log10x
(1,0)
-f(x) ( reflect in x - axis )
y = log10x
y = -log10x
(10,1)
(10,-1)
Graph Sketching
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Higher
Outcome 3
Here is the graph of
y = ex
y = ex
(1,e)
(0,1)
Sketch the graph of
y = -e(x+1)
Graph Sketching
Outcome 3
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Higher
If f(x) = ex
-e(x+1) = -f(x+1)
reflect in x-axis
move 1 left
(-1,1)
y = -e(x+1)
(0,-e)
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Revision
Logarithms & Exponentials
Higher Mathematics
Next
Logarithms
Revision
Reminder
All the questions on this topic will depend
upon you knowing and being able to use,
some very basic rules and facts.
When you see this button
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show
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Logarithms
Revision
Three Rules of logs
log a x  log a y  log a xy
x
log a x  log a y  log a
y
log a x  p log a x
p
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Logarithms
Revision
Two special logarithms
log a a  1
log a 1  0
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Logarithms
Revision
Relationship between log and exponential
log a x  y  a  x
y
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Logarithms
Revision
Graph of the exponential function
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Logarithms
Revision
Graph of the logarithmic function
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Logarithms
Revision
Related functions of
y  f ( x)
y  f ( x  a)
Move graph left a units
y  f ( x  a)
Move graph right a units
y   f ( x)
Reflect in x axis
y  f ( x)
Reflect in y axis
y  f ( x)  a
Move graph up a units
Move graph down a units
y  f ( x)  a
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show
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Logarithms
Revision
Calculator keys
Back
ln
=
log e
log
=
log10
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Logarithms
Revision
Calculator keys
loge 2.5
=
ln
2
.
5
=
= 0.916…
log10 7.6
=
log
7
.
6
=
= 0.8808…
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Logarithms
Revision
Solving exponential equations
2.4  3.1e
x
Use log ab = log a + log b
log e 2.4  log e 3.1e x
x
log e 2.4  log e 3.1  log e e
Use log ax = x log a
loge 2.4  loge 3.1  x loge e
Use loga a = 1
log e 2.4  log e 3.1  x
Take loge both sides
x  log e 2.4  log e 3.1  0.25593...  0.26 (2dp)
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Logarithms
Revision
Solving exponential equations
60  80e
Take loge both sides
Use log ab = log a + log b
k
log e 60  log e 80e k
log e 60  log e 80  log e e
k
Use log ax = x log a
loge 60  loge 80  k loge e
Use loga a = 1
loge 60  log e 80  k
k  loge 80  log e 60
Back
 0.2876...
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 0.29 (2dp)
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Logarithms
Revision
Solving logarithmic equations
log3 y  0.5
0.5
y 3
Change to exponential form

y 3
Change to exponential form
1
2

1
3
1
2
1

3
y  0.577....  0.58 (2dp)
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Logarithms
Revision
3loge (2e)  2log e (3e)
expressing your answer in the form A  log e B  log e C
Simplify
where A, B and C are whole numbers.
log e (2e)  log e (3e)
3
 log e 8e3  log e 9e2
2
8e3
log e 2
9e
8e
 log e
9
 loge 8  log e e  log e 9
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 1  loge 8  log e 9
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Logarithms
Revision
Simplify
log5 2  log5 50  log5 4
2  50
 log 5
4
 log 5 52
Back
 log5 25
 2
 2 log5 5
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Logarithms
Revision
Find x if
4 log x 6  2 log x 4  1
64
 log x 2  1
4
 log x 64  log x 42  1
9
 log x
36  36
1
4 4
1
9
 log x 81  1
1
 x  81
 x1  81
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Logarithms
Revision
Given x  log5 3  log5 4 find algebraically the value of x.
 x  log5 3  4
 x  log5 12
 5x  12
 x log10 5  log10 12
 log10 5x  log10 12
log10 12
 x
log10 5
Back
 x  1.5439..
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 x  1.54 (2 dp)
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Logarithms
Revision
Find the x co-ordinate of the point where the graph of the curve
with equation y  log3 ( x  2)  1 intersects the x-axis.
When y = 0
0  log3 ( x  2)  1
Re-arrange
1  log3 ( x  2)
Exponential form
Re-arrange
Back
31  x  2
1
x  23
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x2
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1
3
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The graph illustrates the law
Logarithms
Revision
y  kx n
If the straight line passes through A(0.5, 0)
and B(0, 1). Find the values of k and n.
log 5 y  log 5 kx n
Gradient
log5 y  log5 k  n log5 x
1
0.5
log5 k  1
Y  log5 k  nX
Back

y-intercept
 k 5
n  2 (gradient)
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1
Logarithms
Revision
Before a forest fire was brought under control, the spread of fire was
described by a law of the form A  A e kt
0
where
A0 is the area covered by the fire when it was first detected
and A is the area covered by the fire t hours later.
If it takes one and a half hours for the area of the forest fire to double,
find the value of the constant k.
2 A0  A0 e k 1.5
 loge 2  1.5k
Back
 2  ek 1.5
 loge 2  1.5k loge e
log e 2
 k
 k  0.462..  0.46
1.5
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(2 dp)
Logarithms
Revision
The results of an experiment give rise to the graph shown.
a) Write down the equation of the line in terms of P and Q.
It is given that P  log e p and
Q  log e q
b) Show that p and q satisfy a relationship of the form
p  aqb
stating the values of a and b.
Gradient
0.6
y-intercept
P  0.6Q  1.8
Back
1.8
log e p  log e aq b
loge p  loge a  b loge q
P  log e a  bQ
log e a  1.8  a  e1.8
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b  0.6
 a  6.05
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Logarithms
Revision
The diagram shows part of the graph of
y  logb ( x  a)
.
Determine the values of a and b.
Use (7, 1)
1  logb (7  a)
...(1)
Use (3, 0)
0  logb (3  a)
...(2)
Hence, from (2)
and from (1)
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3 a 1
a  2
b5
1  logb 5
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Logarithms
Revision
The diagram shows a sketch of
part of the graph of y  log 2 ( x)
a) State the values of a and b.
b) Sketch the graph of y  log 2 ( x  1)  3
a 1
b3
Graph moves
1 unit to the left and 3 units down
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Logarithms
Revision
a) i) Sketch the graph of y  a x  1, a  2
ii) On the same diagram, sketch the graph of
y  a x 1 , a  2
b) Prove that the graphs intersect at a point
where the x-coordinate is
 1 
log a 

a

1


a x  1  a x 1
 1  a x1  a x
 loga 1  loga a x  loga  a 1
 x   loga  a  1
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 1  a x  a 1
 0  x  log a  a  1
 x  log a  a  1
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1
 1 
 x  log a 

a

1




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Logarithms
Revision
Part of the graph of
y  5 log10 (2 x  10)
is shown in the diagram.
This graph crosses the x-axis at
the point A and the straight line
y  8 at the point B. Find algebraically the x co-ordinates of A and B.
8  5 log10 (2 x  10) 
 101.6  2 x  10
0  5 log10 (2 x  10)
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8
 1.6  log10 (2 x  10)
 log10 (2 x  10)
5
 101.6  10  2x  x  14.9 B (14.9, 8)
 2 x  10  1
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 x  4.5
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A (4.5, 0)
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Logarithms
Revision
The diagram is a sketch of part of the graph of
y  ax , a  1
a) If (1, t) and (u, 1) lie on this curve, write down
the values of t and u.
b) Make a copy of this diagram and on it sketch
2x
y

a
the graph of
c) Find the co-ordinates of the point of intersection of
y  a 2x with the line x  1
b)
t

a
u

0
a)
c)
y  a 21
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y  a2
1,
a2 
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Logarithms
Revision
The diagram shows part of the graph with equation
y  3x and the straight line with equation y  42
These graphs intersect at P.
Solve algebraically the equation 3x  42
and hence write down, correct to 3 decimal places, the co-ordinates of P.
log10 3x  log10 42
log10 42
 x
log10 3
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 x log10 3  log10 42
 x  3.40217...
 P (3.402, 42)
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