The Exponential and natural log functions

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…


£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 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He introduced the number as a base of logarithms and started to use the letter e when writing an unpublished paper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.



y = ex

y = 2ex

y = ex

(0,1)f(x)

2f(x)

y = 2ex

(0,2)


(For the same set of inputs (x),

the outputs (y) double)


y = ex

y = ex + 2

y = ex

(0,1)f(x)

f(x) + 2

y = ex + 2

(0,3)



the outputs (y) increase by 2)


y = ex

y = -ex

y = ex

(0,1)

f(x)

-f(x)

y = -ex

(0,-1)



the outputs (y) ‘swap signs’


y = ex

y = e2x

y = ex

(0,1)

f(x)

f(2x)

y = e2x


(The same set of outputs (y) for

half the inputs (x))


y = ex

y = ex + 1

y = ex

(0,1)

f(x)

f(x + 1)

y = ex + 1



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


y = ex

y = e-x

y = ex

(0,1)

f(x)

f(-x)

y = e-x



inputs with the opposite sign…

(0,1)


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)



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 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.


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)




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…




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…




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…




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)




y = lnx

(1,0)y = ln(x) f(x)

y = ln(x + 2) f(x + 2)

y = ln(x + 2)


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)




y = lnx

(1,0)y = ln(x) f(x)

y = ln(-x) f(-x)

y = ln(-x)


input (x) values with the opposite sign

to before

(-1,0)




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




y = lnx

(1,0)


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.


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.


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…






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






ln(3 2) 3x ‘Reverse ln’

Add 2

33 2x e 33 2x e 3 2

3

ex

Divide by 3

Example Question 2


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!)



3 22 2 0y y ye e e

Example Question 4

Solve






MIXED EXERCISE Page 75

Question 1 (any 2 parts)






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


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

The Exponential and natural log functions

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