1. The Exponential & Logarithmic Functions
Exponential Graph
Logarithmic Graph
(a , 1)
(1 , 0)
(1 , a)
(0 , 1)
y = ax
y = logax

2. Exponential Growth & Decay
Joan puts £2500 into a savings account earning 13% interest per annum. How much money will she have if she leaves it there for 15 years?
Let £A(n) be the amount in her account after n years, then:
Multiplying factor is 1∙13
A(0) = 2 500
The initial investment
A(1) = × 1∙13
After 1 year
A(15) = × 1∙1315
= £15 635·68
A(2) = × 1∙132
A(3) = × 1∙133
A(n) = × 1∙13n
After n years

3. Multiplying factor is 0∙98
The population of an urban district is decreasing at the rate of 2% per year.
(a) Taking P0 as the initial population, find a formula for the population, Pn, after n years.
(b) How long will it take for the population to drop from to ?
Multiplying factor is 0∙98 P0
The initial population
Pn = , P0 =
P1 = 0∙98×P0
0∙98n× =
P2 = 0∙982×P0
0∙98n× =
And so on
P3 = 0∙983×P0
Takes 5 years to drop below
Pn = 0∙98n×P0
Population after n years

4. A Special Exponential Function – the “Number” e
The letter e represents the value 2·
Exponential functions are used to model populations, carbon date artifacts, help coroners determine time of death, compute investments.
f(x) = 2·718..x = ex is called the exponential function to the base e.

5. What is the mass after 12 years?
The mass of a fixed quantity of radioactive substance decays according to the formula m = 50e-0∙02t, where m is the mass in grams and t is the time in years.
What is the mass after 12 years?
m = 50 e – 0∙02t
m = 50 e – 0∙02×12
m = 50 e – 0∙24
m = 39∙3 grams

6. Linking the Exponential and Logarithmic Function
y = ax
loga 1 = 0
 a0 = 1
y = x
loga a = 1
 a1 = a
loga y = x
 ax = y
(0,1)
(1,0)
The log function and the exponential functions are reflections of each other in the line y = x
y = logax
The log function is the inverse of the exponential function, so it ‘undoes’ or reverses the exponential function:

7. Linking the Exponential and the Log Function
f(x) = 2x
2 to what power gives the required number?
f(1) →
f(2) →
f(3) →
f(4) → 21 22 23 24
so log22 = 1
so log24 =
so log28 =
so log216 = 2 3 4
f(x) = log2x

8. Linking the Exponential and the Log Function
It is useful to remember a basic connection
23 = 8
log2 8 = 3

9. Linking the Exponential and the Log Function
Examples: = 8 ↔ log28 = 3
1) = ↔ 49
log749 = 2
2) log = ↔ 3
103 = 1000
3) log5125 = ↔ 3
53 = 125
4) = ↔
128
log2128 = 7
5) = ↔
243
log3243 = 5

10. Some Rules of Indices

12. Examples : Simplify the following

14. (a) log 381 = “ to what power gives ?”4 3 81
34 = 81
(b) log 42 = “ to what power gives ?”
1/2 4 2
√4 = 2
√4 = 41/ 2
(c) log = “ to what power gives ?”
1/27 -3 3
1/27
1/27 = 1/33
= 3-3

15. logax + logay = logaxy logax – logay = logax/y logaxn = n logax
Rules of Logarithms
logax + logay = logaxy
logax – logay = logax/y
logaxn = n logax

16. Rules of Logarithms Examples
Simplify:
b) log363 – log37
a) log102 + log10500
logax + logay = logaxy
logax – logay = logax/y
= log10 2×500
= log363/7
= log
= log3 9
= log3 (3)2
= log10 (10)3
= 2
= 3

17. logaxn = n logax  n logax = logaxn
Simplify
logaxn = n logax  n logax = logaxn
logax – logay = logax/y
= log2 (2)1

18. You have 2 logarithm buttons on your calculator:
Using your Calculator
You have 2 logarithm buttons on your calculator:
which stands for log10
log
10x
and its inverse
which stands for loge ln
and its inverse ex

19. Using your Calculator
23 = 8
 log2 8 = 3
 23 = 8
1) log10 x = 3∙5
 x = 10 3∙5
 x = 3162
2) ln x = 3∙5
 x = e 3∙5
 x = 33∙1
3) e x = 47
 loge 47 = x
 ln 47 = x
 x = 3∙85
4) 10 x = 78
 log10 78 = x
 x = 1∙89

20. Solving Exponential Equations
Solve 5x = 32
loga xn = n loga x
Take logs of both sides
log 5x = log 32
ln 5x = ln 32
 x log 5 = log 32
 x ln 5 = ln 32
 x = log 32 ÷ log 5
 x = ln 32 ÷ ln 5
 x = 2∙15
 x = 2∙15

21. Solving Exponential Equations
Solve 7x = 90
loga xn = n loga x
Take logs of both sides
log7 x = log 90
ln 7x = ln 90
 x log 7 = log 90
 x ln 7 = ln 90
 x = log 90 ÷ log 7
 x = ln 90 ÷ ln 7
 x = 2∙31
 x = 2∙31

22. Solving Exponential Equations
If the equation involves 10x or ex then we do not need to take logs of both sides.
Just “undo” the power using 23 = 8  log28 = 3
ex = 34
 x = ln 34
= 3∙53
e5p = 187
 5p = ln 187
= 5∙23
 p = 5∙23 ÷ 5 = 1∙05
6e1∙8x = 906
 e 1∙8x = 151
 1∙8x = ln 151
 x = 2∙79

23. 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?
When t = 0 A = A0 = 50g
When t = 4 A = 45g
 A = 50e – kt
 45 = 50e – k×4
 45/50 = e – 4k
= 0∙9
 ln 0∙9 = – 4k
 k = ln 0∙9 ÷ – 4
= 0∙026

24. 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?
When A = 1/2A0 = 25g
 25 = 50e – 0∙026t
 25/50 = e – 0∙026t
= 0∙5
 ln 0∙5 = – 0∙026t
 t = ln 0∙5 ÷ – 0∙026
= 26∙7 mins

25. Experiment and Theory
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

26. We can convert y = kxn using logarithms:
Take logs on both sides of y = kxn
log y = log kxn
log xy = log x + log y
 log y = log xn + log k
log x n = n log x
log y
log x
 log y = n log x + log k
Compare this to y = mx + c a straight line
Gradient n
(0 , log k)
y = kxn is converted into a straight line using logarithms

27. The following data was collected during an experiment:
50∙1
194∙9
501∙2
707∙9 y
20∙9
46∙8
83∙2
102∙3
y and x are related by the formula y = kxn, find the values of k and n and state the formula that connects x and y.
log y = n log x + log k
log 20∙9 = n log 50∙1 + log k
 1∙32 = n×1∙70 + log k
 2∙01 = n×2∙85 + log k
log 102∙3 = n log 707∙9 + log k
 n = 0∙6, log k = 0∙3
 y = 2x 0∙3
 k = 2∙0

28. Variables x and y are related by the equation y = kxn.
Find the values of k and n.
log2 y = log2 kxn
log2 k = 5
 k = 25
= 32
log2 y = log2 xn + log2 k
log2 y = nlog2 x + log2 k
 n = 0∙5

29. N = Noert No = 61 million, r = 1∙6, t = 14
The size of the human population, N, can be modelled using the equation N = N0ert where N0 is the population in 2006, t is the time in years since 2006, and r is the annual rate of increase in the population.
(a) In 2006 the population of the United Kingdom was approximately 61 million, with an annual rate of increase of 1∙6%. Assuming this growth rate remains constant, what would be the population in 2020?
N = Noert
No = 61 million, r = 1∙6, t = 14
N = 61e0∙016  14
= 61e0∙224
= 61× 1∙251
= 76∙3 million

30. No = 5∙1 million, r = 0∙43, N = 10∙2 million
The size of the human population, N, can be modelled using the equation N = N0ert where N0 is the population in 2006, t is the time in years since 2006, and r is the annual rate of increase in the population.
(b) In 2006 the population of Scotland was approximately 5∙1 million, with an annual rate of increase of 0.43%. Assuming this growth rate remains constant, how long would it take for
Scotland’s population to double in size?
N = Noert
No = 5∙1 million, r = 0∙43, N = 10∙2 million
10∙2 = 5∙1e0∙0043 t
 2 = e0∙0043 t
 ln 2 = 0∙0043 t
 t = ln 2 ÷ 0∙0043
 t = 161 years

31. a) 6 = 3  4a  2 = 4a  a = ½ b) b = 3  4-1/2  b = 3  1/√4
Two variables x and y satisfy the equation y = 3 × 4x.
(a) Find the value of a if (a, 6) lies on y = 3 × 4x.
(b) If (-1/2, b ) also lies on the graph, find b.
(c) A graph is drawn of log10 y against x. Show that its equation will be of the form log10 y = Px + Q and state the gradient of this line.
a) 6 = 3  4a
 2 = 4a
 a = ½
b) b = 3  4-1/2
 b = 3  1/√4
 b = 11/2
c) log10 y = log103  4x
log10 y = log103 + log104x
log10 y = log103 + xlog104
Gradient = log104

32. A(t) = 0∙88Ao 0∙88Ao = Aoe -0∙000124t 0∙88 = e -0∙000124t
It is claimed that a wheel made from wood is over 1000 years old.
Carbon dating is used to test the claim, using the formula A(t) = Aoe - 0∙000124t to determine the age of the wood. Ao is the amount of carbon in any living tree, A(t) is the amount of carbon after t years. For the wheel it was found that A(t) was 88% of the amount of carbon in a living tree. Is the claim true?
A(t) = 0∙88Ao
0∙88Ao = Aoe -0∙000124t
0∙88 = e -0∙000124t
loge0∙88 = – 0∙000124t
t = ln 0∙88 ÷ – 0∙000124
t = 1031
Claim true

33. This is the graph of y = 2mx
What is the value of m?
(3 , 54) on curve  54 = 2m3
 27 = m3
 m = 3

34. What is the value of q?
(q , 2) on curve  2 = log3(q – 4)
 23 = q – 4
 8 = q – 4
 q = 12

35. You must know and use these rules in both directions
Rules of
Logarithms
loga x + loga y  loga xy
You must know and use these rules in both directions
loga x – loga y  loga x/y
loga xn  n loga x
loga 1 = 0
loga a = 1

36. Linking the Exponential and Logarithmic Function
y = ax
loga 1 = 0
 a0 = 1
y = x
loga a = 1
 a1 = a
loga y = x
 ax = y
(0,1)
(1,0)
The log function and the exponential functions are reflections of each other in the line y = x
y = logax
The log function is the inverse of the exponential function, so it ‘undoes’ or reverses the exponential function:

37. Simplify log5 2 + log5 50 – log5 4
loga x – loga y = loga x/y
loga x + loga y = loga xy
52 = 25

38. Find x if logx 6 – 2 logx 4 = 1
loga x – loga y = loga x/y
n loga x = loga xn
4 logx 6 – 2 logx 4 = 1

39. Given x = log5 3 + log5 4 find algebraically the value of x.
Take logs on both sides
log10 works just as well

40. The graph illustrates the law y = kxn
If the straight line passes through A(0∙5, 0) and B(0, 1). Find the values of k and n.
log5y = log5kxn
log5y = log5k + log5xn
log5k =1
 k = 51
 k = 5
log5y = log5k + n log5x
log5y = n log5x + log5k
n = –1/0∙5
y = mx + c
 n = – 2

41. Before a forest fire was brought under control, the spread of fire was described by a law of the form
where A0is 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.

42. b) Show that p and q satisfy p = aqb
The results of an experiment give rise to the graph shown.
Write down the equation of the line in terms of P and Q given that
P = logep and Q = logeq
b) Show that p and q satisfy p = aqb
Gradient
y-intercept

43. Hence, from (2) and from (1)
The diagram shows part of the graph of
Determine the values of a and b.
Use (7, 1)
Use (3, 0)
Hence, from (2)
and from (1)

44. a) i) Sketch the graph of
ii) On the same diagram, sketch
Prove that the graphs intersect at a point where the x-coordinate is