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Introduction
• We are going to look at exponential functions
• We will learn about a new ‘special’ number in Mathematics
• We will see how this number can be used in practical problems…
The Exponential and Log Functions
3A
Imagine you have £100 in a bank account
Imagine your interest rate for the year is 100%
You will receive 100% interest in one lump at the end of the year, so you will now have £200 in the bank
However, you are offered a possible alternative way of being paid
Your bank manager says, ‘If you like, you can have your 100% interest split into two 50% payments, one made halfway through the year, and one made at the end’
How much money will you have at the end of the year, doing it this way (and what would be the quickest calculation to work that out?)
£100 x 1.52
= £225
Investigate further. What would happen if you split the interest into 4, or 10, or 100 smaller bits etc…
The Exponential and Log Functions
3A
£100e£100 x (1 + 1/n)n100/nn£100
£271.81£100 x 1.0001100000.01%10,000£100
£271.69£100 x 1.00110000.1%1,000£100
£270.48£100 x 1.011001%100£100
£269.16£100 x 1.02502%50£100
£265.33£100 x 1.05205%20£100
£259.37£100 x 1.11010%10£100
£256.58£100 x 1.125812.5%8£100
£244.14£100 x 1.25425%4£100
£225£100 x 1.5250%2£100
£200£100 x 2100%1£100
Total (2dp)SumInterest Each
PaymentPayments
Start Amount
11
n
en
The larger the value of n, the better the accuracy of e…The value of e is irrational, like π…It also has another interesting property…
(2.718281828459…)
The Exponential and Log Functions
3A
Gradient Functions
You have already learnt about differentiation, that differentiating a graph function gives the gradient function…
We can plot the gradient function on the graph itself…
y = x2
So dy/dx = 2x
y = x2
y = 2x
The Gradient at this point…
… is this value here!
The Exponential and Log Functions
3A
Gradient Functions
You have already learnt about differentiation, that differentiating a graph function gives the gradient function…
We can plot the gradient function on the graph itself…
y = x3
So dy/dx = 3x2
y = x3y = 3x2
The Gradient at this point…
… is this value here!
And the Gradient is the same
here!
The Exponential and Log Functions
3A
Gradient Functions
You have already learnt about differentiation, that differentiating a graph function gives the gradient function…
We can plot the gradient function on the graph itself…
y = 2x
dy/dx = 2xln2
At this stage, you do not need to know where this comes
from…
y = 2x
y = 2xln2
The Exponential and Log Functions
3A
Gradient Functions
You have already learnt about differentiation, that differentiating a graph function gives the gradient function…
We can plot the gradient function on the graph itself…
y = 3x
dy/dx = 3xln3
At this stage, you do not need to know where this comes
from…
y = 3x
y = 3xln3
The Exponential and Log Functions
3A
y = 3xy = 3xln3
y = 2x y = 2xln2
What has happened from the first graph to the second?
The lines have crossed…
Therefore the must be a value between 2 and 3 where the lines are equal…
The Exponential and Log Functions
3A
Gradient Functions
If we plot a graph of ex, its gradient function is the same graph!
This leads to an interesting conclusion…
If y = ex
Then dy/dx = ex as well!
y = ex
y = ex
The Exponential and Log Functions
• The Graph of ex is used rather than 2x or 3x because e is the increase when growth is continuous (ie growth is always happening, rather than in ‘chunks’)
• The majority of growth follows this format, bacterial growth is continuous for example.
• Transforming the graph can also allow graphs to show decreases, such as radioactive decay and depreciation
3A
You need to be able to sketch transformations of the graph y = ex
y = ex
y = 2ex
3A
y = ex
(0,1)f(x)
2f(x)
y = 2ex
(0,2)
The Exponential and Log Functions
(For the same set of inputs (x), the
outputs (y) double)
You need to be able to sketch transformations of the graph y = ex
y = ex
y = ex + 2
3A
y = ex
(0,1)f(x)
f(x) + 2
y = ex + 2
(0,3)
The Exponential and Log Functions
(For the same set of inputs (x), the
outputs (y) increase by 2)
You need to be able to sketch transformations of the graph y = ex
y = ex
y = -ex
3A
y = ex
(0,1)
f(x)
-f(x)
y = -ex
(0,-1)
The Exponential and Log Functions
(For the same set of inputs (x), the outputs (y) ‘swap
signs’
You need to be able to sketch transformations of the graph y = ex
y = ex
y = e2x
3A
y = ex
(0,1)
f(x)
f(2x)
y = e2x
The Exponential and Log Functions
(The same set of outputs (y) for half
the inputs (x))
You need to be able to sketch transformations of the graph y = ex
y = ex
y = ex + 1
3A
y = ex
(0,1)
f(x)
f(x + 1)
y = ex + 1
The Exponential and Log Functions
(The same set of outputs (y) for
inputs (x) one less than before…)
(0,e)
We can work out the y-intercept by substituting in x = 0
This gives us e1 = e
You need to be able to sketch transformations of the graph y = ex
y = ex
y = e-x
3A
y = ex
(0,1)
f(x)
f(-x)
y = e-x
The Exponential and Log Functions
(The same set of outputs (y) for inputs with the opposite sign…
(0,1)
You need to be able to sketch transformations of the graph y = ex
Sketch the graph of:
y = 10e-x
3A
y = ex
The graph of e-x, but with y values 10 times bigger…
y = e-
x
The Exponential and Log Functionsy =
10e-x
(0, 1)
(0, 10)
You need to be able to sketch transformations of the graph y = ex
Sketch the graph of:
y = 3 + 4e0.5x
3A
y = ex
The graph of e0.5x, but with y values 4 times bigger with 3
added on at the end…
(0, 1)
(0, 7)
y = e0.5x
y = 4e0.5x
y = 3 + 4e0.5x (0, 4)
The Exponential and Log Functions
The Exponential and Log Functions
You need to be able to used the Exponential and Log
Functions to solve problems…
The Price of a used car is given by the formula:
a) Calculate the value of the car when it is new The new price implies t, the time, is 0… Substitute t = 0 into the formula…
3A
P = 16000e
- t 10
P = 16000e
- t 10
P = 16000e
- 0 10
P = 16000e
0
P = £16000
The Exponential and Log Functions
You need to be able to used the Exponential and Log
Functions to solve problems…
The Price of a used car is given by the formula:
b) Calculate the value after 5 years… 5 years implies t = 5 Substitute t = 5 into the formula…
3A
P = 16000e
- t 10
P = 16000e
- t 10
P = 16000e
- 5 10
P = 16000e
-0.5
P = £9704.49
The Exponential and Log Functions
You need to be able to used the Exponential and Log
Functions to solve problems…
The Price of a used car is given by the formula:
c) What is the implied value of the car in the long run (ie – what value does it tend towards?) Imagine t tends towards infinity (gets really big)
3A
P = 16000e
- t 10
P = 16000e
- t 10
P = 16000 x 0
P = £0
e- t
101
(10√e)t
Bigger t
= Bigger denominator
= Smaller Fraction value…
The Exponential and Log Functions
You need to be able to used the Exponential and Log
Functions to solve problems…
The Price of a used car is given by the formula:
d) Sketch the Graph of P against t Value starts at £16000 Tends towards 0, but doesn’t get there…
3A
P = 16000e
- t 10
P = 16000e
- t 10
P
t
£16000
t is independent so goes on the x axis
P is dependant on t so goes on the y axis
The Exponential and Log Functions
You need to be able to solve equations
involving natural logarithms and e
This is largely done in the same way as in C2 logarithms, but using ‘ln’ instead of ‘log’
3B
Example Question 1
3xe
ln( ) ln(3)xe
ln( ) ln(3)x e
ln(3)x
1.099x
Take natural logs of both sides
Use the ‘power’ lawln(e) = 1
(e to the power something is e)
Work out the answer or leave as a logarithm
You do not necessarily need to write these
steps…
The Exponential and Log Functions
You need to be able to solve equations
involving natural logarithms and e
This is largely done in the same way as in C2 logarithms, but using ‘ln’ instead of ‘log’
3B
Example Question 1
2 7xe Take natural logs
Use the power law
2ln( ) ln(7)xe
2 ln(7)x
ln(7) 2x
0.054x
Subtract 2
Work out the answer or leave as a logarithm
The Exponential and Log Functions
You need to be able to solve equations
involving natural logarithms and e
This is largely done in the same way as in C2 logarithms, but using ‘ln’ instead of ‘log’
3B
Example Question 1
ln 4x ‘Reverse
ln’Work it out if
needed…
4x e
54.598x
The Exponential and Log Functions
You need to be able to solve equations
involving natural logarithms and e
This is largely done in the same way as in C2 logarithms, but using ‘ln’ instead of ‘log’
3B
Example Question 1
ln(3 2) 3x ‘Reverse
ln’
Add 2
33 2x e 33 2x e
3 2
3
ex
Divide by 3
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
The inverse of ex is logex (usually written as lnx)
We know from chapter 2 that an inverse function is a
reflection in the line y = x…
3B
y = ex
y = lnx
y = x
(0,1)
(1,0)
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
3B
y = lnx
(1,0)
y = lnx f(x)
y = 2lnx 2f(x)
y = 2lnx
All output (y) values doubled for the same
input (x) values…
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
3B
y = lnx
(1,0)
y = lnx f(x)
y = lnx + 2
f(x) + 2
y = lnx + 2
(0.14,0)
ln 2y x
0 ln 2x
2 ln x 2e x
0.13533... x
Let y = 0
Subtract 2
Inverse ln
Work out x!
All output (y) values increased by 2 for the
same input (x) values…
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
3B
y = lnx
(1,0)y = lnx f(x)
y = -lnx -f(x)y = -lnx
All output (y) values ‘swap sign’ for the
same input (x) values…
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
3B
y = lnx
(1,0)y = ln(x) f(x)
y = ln(2x) f(2x)
y = ln(2x)
All output (y) values the same, but for half the input (x) values…
(0.5,0)
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
3B
y = lnx
(1,0)y = ln(x) f(x)
y = ln(x + 2)
f(x + 2)
y = ln(x + 2)
All output (y) values the same, but for
input (x) values 2 less than before
(-1,0)
ln( 2)y x
ln(2)y
0.69314...y
Let x = 0
Work it out (or leave as
ln2)
(0, ln2)
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
3B
y = lnx
(1,0)y = ln(x) f(x)
y = ln(-x) f(-x)
y = ln(-x)
All output (y) values the same, but for
input (x) values with the opposite sign to
before
(-1,0)
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
3B
y = lnx
(1,0)
Sketch the graph of:
y = 3 + ln(2x)
y = ln(2x)
(0.025,0)
y = 3 + ln(2x)
The graph of ln(2x), moved up 3
spaces…
3 ln(2 )y x 0 3 ln(2 )x 3 ln(2 )x 3 2e x 3
2
ex
Let y = 0
Subtract 3
Reverse ln
Divide by 2
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
3B
y = lnx
(1,0)
Sketch the graph of:
y = ln(3 - x)
The graph of ln(x), moved left 3 spaces, then
reflected in the y axis. You must do the reflection last!
y = ln(3 + x)
(-2,0)
(2,0)
y = ln(3 - x)
ln(3 )y x
ln(3)y Let x = 0
(0,ln3)
x = 3
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
The number of elephants in a herd can be represented by the equation:
Where n is the number of elephants and t is the time in years after
2003.
a) Calculate the number of elephants in the herd in 2003 Implies t = 0
3B
40150 80t
N e
40150 80t
N e
0
40150 80N e
0150 80N e
150 80N
70N
t = 0
e0 = 1
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
The number of elephants in a herd can be represented by the equation:
Where n is the number of elephants and t is the time in years after
2003.
b) Calculate the number of elephants in the herd in 2007 Implies t = 4
3B
40150 80t
N e
40150 80t
N e
4
40150 80N e
150 72.4N
77.6 (78)N
t = 4
Round to the nearest whole
number
The Exponential and Log Functions
You need to be able to plot and understand
graphs of the function which is inverse to ex
The number of elephants in a herd can be represented by the equation:
Where n is the number of elephants and t is the time in years after
2003.
c) Calculate the year when the population will first exceed 100
elephants Implies N = 100
3B
40150 80t
N e
40150 80t
N e
40100 150 80t
e
N = 100
4050 80t
e
400.625t
e
ln(0.625)40
t
40ln(0.625) t
18.8 (19 years)t
2003 + 19 = 2022
Subtract 150
Divide by -80
Take natural logs
Multiply by 40
The Exponential and Log Functions
You need to be able to plot and understand
graphs of the function which is inverse to ex
The number of elephants in a herd can be represented by the equation:
Where n is the number of elephants and t is the time in years after
2003.
d) What is the implied maximum number in the herd?
Implies t ∞
3B
40150 80t
N e
40
t
e
40150 80t
N e
40
1t
e
Rearrange
As t increases
Denomintor becomes bigger
Fraction becomes smaller, towards 0
150N
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
f(x) = x – 1g(x) = ex + 1
a) Calculate gf(4)
3B
The Exponential and Log Functions
You need to be able to plot and understand graphs of
the function which is inverse to ex
f(x) = x – 1g(x) = ex + 1
b) Calculate g-1(x)
3B
Summary• We have learnt a new number, e, and seen what is
stands for and where it is used
• We have plotted the graph of ex and lnx, which are inverse functions
• We have seen how to transform these graphs
• We have solved practical problems involving theese
• We have also seen again how to solve logarithmic equations…