the exponential and natural log functions
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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
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
£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 PaymentPaymentsStart Amount
11
n
en
The larger the value of n, the better the accuracy of e…(2.718281828459…)
The Exponential and Log Functions
The mathematical constant e was invented by a Scottish scientist John Napier and was first used by a Swiss
Mathematician Leonhard Euler.
He introduced the number as a base of logarithms and started to use the letter e when writing an unpublished paper
on explosive forces in Cannons.
e, (Euler’s constant) is an irrational number whose value is e = 2.718281…
When mathematicians talk about exponential functions they are referring to function ex, where e is the constant.
Graph of ex
The following diagram shows graph of y = ex
It is also known as an exponential graph.
You need to be able to sketch transformations of the graph y = ex
So lets recap our transformations
TASK 1: Match the equations to the graphs.
TASK 2: Fill in the table. Describe what transformation is taking place.
Is it affecting the x or y?Change the coordinates after the transformation.
The Exponential and Log Functions
You need to be able to sketch transformations of the graph y = ex
y = ex
y = 2ex
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
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
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
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
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
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
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
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 Logarithmic Function
e is often used as a base for a logarithm.
This logarithm is called the natural logarithm
where logex is written as: ln x
Graph of ln xThe following diagram shows graph of y = ln x.
Note that the graph of ln x does not
exist in the negative x-axis.
So, ln x does not exist for negative
value of x.
Try it on your calculator now.
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)
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
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
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
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
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
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
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
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
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)
Describe the transformations of the lnx and ex functions.
Draw the old and new curves (on the same graph) for each question.
The Exponential and Log Functions
What’s the connection?
You already know that every logarithmic function
has an inverse involving an exponential function.
e.g. log2x = 5 x = 25
Hence, the natural log, ln x has an inverse of
the exponential function ex.
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’
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’ law
ln(e) = 1
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’
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
TRY THIS ONE
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’
ln(3 2) 3x ‘Reverse ln’
Add 2
33 2x e 33 2x e 3 2
3
ex
Divide by 3
Example Question 2
The Exponential and Log Functions
You need to be able to solve polynomials involving natural logarithms and e
2(ln ) 3ln 2 0x x
Example Question 3
Solve
These equations always look scary,
however they are only disguised cubic or
quadratic equations (which you can solve!)
The Exponential and Log Functions
You need to be able to solve polynomials involving natural logarithms and e
3 22 2 0y y ye e e
Example Question 4
Solve
These equations always look scary,
however they are only disguised cubic or
quadratic equations (which you can solve!)
The Exponential and Log Functions
You need to be able to solve polynomials involving natural logarithms and e
MIXED EXERCISE Page 75
Question 1 (any 2 parts)
Question 2 (any 3 parts)
Question 3 (any 3 parts)
These equations always look scary,
however they are only disguised cubic or
quadratic equations (which you can solve!)
Differentiation of ex
ex is the only function which is
unchanged when differentiated.
i.e.
xy exdye
dx
Differentiation of ex
xy e xdye
dx
kxy e kxdyke
dx
The Exponential and Log Functions
You need to be able to differentiate
exponential functions
Examples
a. y = e-3x
b. y = x5 – e4x
c. y = e2x + 3
d. y = 5
2
5x
x
e
e
Exercise 4A Page 64
Question 3, 5, 6, 7
Differentiation of ln x
Using the exponential function ex it can be
proved that the differentiation of ln x is
1dy
dx xlny x
Proof of Differentiation of ln xConsider the function y = ln x y = loge x
In exponential form: x = ey
Consider, as a differentiation of ln x with respect to x.
As, x = ey
This means, Proved.
dy
dx
dxx
dy
1dy
xdx
Examples
Find for each of the following:
a. y = x2 – ln(3x)
b. y = 5x3 – 6lnx + 1
Exercise 4E Page 72
Question 1, 7 and 9
Integration of ex
xy e c xdye
dx
1 kxy e ck
kxdye
dx
lny x c 1dy
dx x
Integration of lnx
Examples
Integrate the following:
a) y = e3x
b) y = e½x
c) 3 2x
yx
Exercise 4B Page 66
Question 4, 5, 6
Exercise 4D Page 70
Question 6, 7
Exercise 4E Page 73
Question 2, 6, 8
Mixed Exercise
EXAM QUESTIONS
Page 76
Question 6, 7, 8, 10
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