EXPONENTIAL AND LOGARITHMIC FUNCTIONS

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EXPONENTIALANDLOGARITHMICFUNCTIONS
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Revision Indices
Examples
a)
34 3 3 3 3 81
b)
Solve 2n 1024
Try 28 256
(too low)
(too low)
29 512
?
210 1024
? n 10
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c)
Find the smallest integer value of n such that 3n
gt 10 000.
Try 37 2187
( lt 10 000)
38 6561
( lt 10 000)
39 19683
( gt 10 000)
? n 9
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Graphs of ax Exponential Functions
y ax
For y ax
if a gt 1
The graph

• Is always positive

• Never crosses the x-axis

• Is increasing

(0,1)

• Passes through (0,1)

• Is called a growth function

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For y ax
if 0 lt a lt 1
y ax
The graph

• Is always positive

• Never crosses the x-axis

• Is decreasing

• Passes through (0,1)

(0,1)

• Is called a decay function

Notes
y ax
- is called an exponential function
- has base a
- a ? 0
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Examples
a)
If you deposit 100 at the Northern Rock Bank,
each year the deposit will grow by 15 (compound
interest).
i.
Find a formula for the amount in the bank after n
years.
ii.
After how many years will the balance exceed 500?
i.
After year 0
y0
100
After year 1
y1
y0 1.15
115
After year 2
y2
y1 1.15
(1.15)2 100
132.25
After year 3
y3
y2 1.15
(1.15)3 100
152.0875
?Formula
yn
(1.15)n 100
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ii.
After how many years will the balance exceed 500?
Try n 9
y9
(1.15)9 100
351.7876292
Try n 10
y10
(1.15)10 100
404.5557736
Try n 11
y11
(1.15)11 100
465.2391396
Try n 12
y12
(1.15)12 100
535.0250105
? The balance will exceed 500 after 12 years.
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b)
The rabbit population on an island increases by
12 per year.
i.
Find a formula for the number of rabbits on the
island after n years.
ii.
How many years will it take for the population to
at least double?
i.
r0
After year 0
After year 1
r1
r0 1.12
After year 2
r2
r1 1.12
(1.12)2 r0
? Formula
rn
(1.12)n r0
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For the population to at least double
ii.
(1.12)n
gt
2
(too low)
Try
(1.12)3
1.404928
(too low)
Try
(1.12)4
1.57351936
(too low)
Try
(1.12)5
1.762341683
(too low)
Try
(1.12)6
1.973822685
Try
(1.12)7
2.2106841407
( gt 2 )
? Population will double within 7 years.
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3. Exponential Functions
If we plot f (x) and f (x) for f (x) 2x
If we plot f (x) and f (x) for f (x) 3x
f (x)
f (x)
f (x)
f (x)
(0,1)
(0,1)
We can see that f (x) is below f (x).
We can see that f (x) is above f (x).
? For some value between 2 and 3, f (x) f (x).
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If we plot f (x) and f (x) for f (x) 2.7x
If we plot f (x) and f (x) for f (x) 2.8x
f (x)
f (x)
f (x)
f (x)
(0,1)
(0,1)
We can see that f (x) is just below f (x).
We can see that f (x) is just above f (x).
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From these graphs we can see that

• If f(x) is an exponential function then f (x) is
  also an exponential function

• f (x) k f (x)

• A value exists between 2.7 and 2.8 for which f
  (x) f (x)

f (x) f (x) ex
This value is
2.718..
This value is denoted by
e
If f (x) ex
then
f (x) ex
(0,1)
f (x) 2.718x ex
is called the exponential function to the base e
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Examples
a)
Evaluate e5.
b)
Solve for x, ex 5.
x 1.609437912
e5 148.4131591
1.61 (2dp)
148.4 (1dp)
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Logarithms
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This is pronounced as log to the base 2 of x.
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Examples
a)
Write the following in logarithmic form
i)
43 64
ii)
25 32
iii)
p 82
(Check 2 2 2 8)
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Laws of Logarithms
Law A
log a a n n
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Law B
Law C
Law D
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Logarithmic Equations
Rules learned can now be used to solve
logarithmic equations.
log 3 x log 3 x3 (log 3 8 log 3 x) 0
log 3 x log 3 x3 log 3 8 - log 3 x 0
?
log 3 x3 log 3 8 0
?
log 3 x3 log 3 8
?
x3 8
?
x 2
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2)
log 4 x 2 log 4 3 0
3)
32x-7 243
35 243
log 4 x log 4 32 0
?
log 4 x - log 4 32
5 2x - 7
?
log 4 x log 4 3-2
2x - 7 5
?
x 3 -2
2x 12
?
x 1/32
x 6
?
x 1/9
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Natural Logarithms
Logarithms to the base e are called natural
logarithms.
log e x ln x
i.e.
log 2.718 x ln x
Examples
Solve the following, rounding correct to 1 d.p.
a)
ln x 5
b)
e x 7
(Take natural logs of each side)
log e x 5
ln e x ln 7
xloge e ln 7
x e 5
x 148.41316
x ln 7
x 148.4 (1dp)
x 1.9459101
x 1.9 (1dp)
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c)
37x2 30
d)
For the formula P(t) 50 e -2t
ln 3 7x2 ln 30
i)
Evaluate P(0).
(7x2) ln 3 ln 30
ii)
For what value of t is P(t) ½ P(0)?
7x2 ln 30/ln 3
i)
P(0) 50 e 2(0)
7x2 3.0959033
50 e 0
7x 1.0959033
50
x 0.1565576
ii)
½ P(0) 25
?
50 e -2t 25
x 0.2 (1dp)
e -2t ½
Remember ln e loge e
ln e -2t ln ½
?
-2t -0.6931471
t 0.3465735
t 0.3 (1dp)
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Formula from experimental data

• If data from an experiment is analysed, say x and
  y, and plotted, and it is found to form a
  straight line then x and y are related by the
  formula

y mx c

• If data from an experiment is analysed, say x and
  y, and plotted, and it is found to form an
  exponential growth curve then x and y are related
  by the formula

y k x n
(k and n are constants)
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Using logarithms y k x n can be expressed as a
straight line.
y k x n
log 10 y log 10 k x n
log 10 y log 10 k log 10 x n
log 10 y log 10 k n log 10 x
log 10 y n log 10 x log 10 k
Y

mX

C
Where
Y log 10 y
? If
y k x n
X log 10 x
C log 10 k
Then
log 10 y n log 10 x log 10 k
n gradient
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When / why do we use this?

• If we are given two variables of data (x and y)
  related in some way and when plotted they produce
  an experimental growth curve, through (0,0), then
  the previously given formula can be used to work
  out k and n. This can then be used to decide the
  formula to relate them (y k x n).

• If we are given two variables of data in
  logarithmic form (log 10 x and log 10 y) and when
  plotted a straight line is produced then it can
  be said that the original data (x and y) would
  produce an exponential growth curve. If it passes
  through (0,0) the previously given formula can be
  used to calculate k and n and hence find the
  formula relating them (y k x n).

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Example
Experimental data are given in the table
2.4
5.62
18.2
31.6
129
x
y
8.3
26.3
115
209
1100
i) Show that x and y are related in some way.
ii) Find the values of k and n (to 1dp) and state
the formula that connects x and y.
log10 x
0.38
0.75
1.26
1.50
2.11
i)
log10 y
0.92
1.42
2.06
2.32
3.04
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As a straight line is produced when log 10 x is
plotted against log 10 y then the formula
relating x and y is of the form y k x n.
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ii)
Take 2 points on best fitting straight
line(0.38,0.92) (2.11,3.04)
Put them into Y nX C, where Ylog10y,
Xlog10x, Clog10k.
0.92 0.38n C
Substitute 1.225 into 1
1
3.04 2.11n C
2
0.92 0.38(1.225) C
0.92 0.46566 C
2
- 1
? C 0.4543352601
2.12 1.73n
C log 10 k
? n 1.225433526
?
log 10 k 0.45433
? n 1.2 (1dp)
?
k 10 0.45433
?
k 2.846657779
?
k 2.8 (1dp)
? Formula relating x and y is
y k x n
y 2.8 x 1.2
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Steps
i) Make a table of log10x and log10y.
ii) Plot log10x against log10y.
iii) If a straight line is produced then state
that the form is y k x n.
iv) Choose 2 points from the best fit line (2 far
apart are best).
v) Substitute 2 points into Y nX C to form 2
equations.
vi) Solve by simultaneous equations.
vii) Value for n can be used in final formula.
viii) Value for C is used to calculate k (i.e. C
log10k ? k 10C).
ix) State complete formula y k x n.
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Steps Version 2
i) Make a table of log10x and log10y.
ii) Plot log10x against log10y.
iii) If a straight line is produced then state
that the form is y k x n.
iv) Choose 2 points from the best fit line (2 far
apart are best).
v) Use these two points to find the gradient.
This will give you n.
vi) Use one of the points and y b m(x a)
to find the equation of the line.
vii) You now have the value of C.
viii) Use the value for C to calculate k (i.e. C
log10k ? k 10C).
ix) State complete formula y k x n.
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Related Graphs
The rules previously learned for graph
manipulation can be applied to logarithmic and
exponential curves.
As ex and ln x are opposite operations (inverse)
then the graph of y ln x is a reflection of y
ex about y x.
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Examples
a)
Sketch the graph of y log 2 x.
i)
On the same diagram sketch y log 2 1/x .
y log2x
y log 2 1/x
y log 2 x
y log 2 x -1
? x 2 y
y -log 2 x
If y 0,
x 1
Graph is a reflection of y log 2 x about the
x-axis.
If y 1,
x 2
( 1 , 0 ) , ( 2 , 1 )
y log21/x
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ii)
Sketch the graph of y log 2 x again.
On the same diagram sketch y log 2 2x .
y log 2 2x
? y log 2 2 log 2 x
y log22x
y 1 log 2 x
y log2x
Graph of y log 2 x moves up 1 place.
Crosses x-axis when y0
? 1 log 2 x 0
log 2 x -1
? x 2 -1
? x ½.
( ½ , 0)
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Further Formula from Experimental Data

• If data from an experiment is analysed, say x and
  y, and plotted, and it is found to form an
  exponential growth curve then x and y are related
  by the formula

y a b x
(a and b are constants)
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Using logarithms y a b x can be expressed as a
straight line.
y a b x
log 10 y log 10 a b x
log 10 y log 10 a log 10 b x
log 10 y log 10 a x log 10 b
log 10 y x log 10 b log 10 a
Y

mx

c
Where
Y log 10 y
? If
y a b x
x x
c log 10 a
Then
log 10 y x log 10 b log 10 a
m log 10 a
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When / why do we use this?

• If we are given two variables of data (x and y)
  related in some way and when plotted they produce
  an experimental growth curve, through (0,?),
  then the previously given formula can be used to
  work out a and b. This can then be used to decide
  the formula to relate them (y a b x).

• If we are given two variables of data in
  logarithmic form (x and log 10 y) and when
  plotted a straight line is produced then it can
  be said that the original data (x and y) would
  produce an exponential growth curve. If it passes
  through (0,?) the previously given formula can
  be used to calculate a and b and hence find the
  formula relating them (y a b x).

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Example
Experimental data are given in the table
2.15
2.13
2.00
1.98
1.95
1.93
x
y
83.33
79.93
64.89
62.24
59.70
57.26
i) Show that x and y are related by the formula y
a b x.
ii) Find the values of a and b (to 1dp) and state
the formula that connects x and y.
x
2.15
2.13
2.00
1.98
1.95
1.93
i)
log10 y
1.921
1.903
1.812
1.794
1.776
1.758
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As a straight line is produced when x is plotted
against log 10 y then the formula relating x and
y is of the form y a b x.
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ii)
Take 2 points on best fitting straight
line(2.15,1.921) (1.93,1.758)
Put them into Y mx c, where Ylog10y, xx, m
log10b, clog10a.
1.921 2.15m c
Substitute m 0.740909 into 1
1
1.758 1.93m c
-1
2
1.921 2.15(0.740909) c
1.921 2.15m c
1.921 1.5929545 c
1
-1.758 -1.93m - c
3
? c 0.3280454
0.163 0.22m
? m 0.740909
m log 10 b
?
log 10 b 0.740909
?
b 10 0.740909
?
b 5.5069241
?
b 5.5 (1dp)
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c log 10 a
?
log 10 a 0.3280454
?
a 10 0.3280454
?
a 2.1283618
?
a 2.1 (1dp)
? Formula relating x and y is
y a b x
y 2.1 (5.5) x
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Steps
i) Make a table of x and log10y.
ii) Plot x against log10y.
iii) If a straight line is produced then state
that the form is y a b x.
iv) Choose 2 points from the best fit line (2 far
apart are best).
v) Substitute 2 points into Y mx c to form 2
equations.
vi) Solve by simultaneous equations.
vii) Value for m is used to calculate b (i.e. m
log10b ? b 10m).
viii) Value for c is used to calculate a (i.e. c
log10a ? a 10c).
ix) State complete formula y a b x.