The exponential function e A reminder from C2. Use your calculator (if you need to) to work out the...

The exponential functione

A reminder from C2.

Use your calculator (if you need to) to work out the following.Why are these the answers? (Hint: What does logax actually mean?)

Question Answer Reason

log39 =

log5125 =

log216 =

log61 =

log648 =

log51/25 =

2

3

4

0

½

-2

32 = 9

53 = 125

24 = 16

60 = 1

641/2 (= √64) = 8

5-2 = 1/25

Follow-up question:Based on your answers to the above, what is an alternative way of writing the statement:

y = logex ?


What you have in fact just done is made x the subject of y = ln x.

i.e. y = ln x x = ey

If we now interchange y and x we obtain y = ex.

What effect does this have on the graph of y = ln x?

y

x

y = ln x

× (1, 0)

× (0, 1)


× (2, ln2)

y = x

× (ln2, 2)

× (3, ln3)

× (ln3, 3)

Interchanging x and y has the effect of reflecting the graph in the line y = x.

y

x

y = ln x

× (1, 0)

× (0, 1)


× (2, ln2)

y = x

× (ln2, 2)

× (3, ln3)

× (ln3, 3)

y = ex

The function y = ex is called the exponential function (more on this later).


You know from your work on functions that reflecting in the line y = x gives an inverse function, so it follows that y = ex and y = ln x are the inverse of each other.

What does this mean?

Investigate, using your calculator, eln x and ln(ex) for different values of x. What is happening?

These functions “undo” each other.

ln(ex) = xlne (using the laws of logarithms) = x (using the fact that ln e = logee = 1.

More on this later!


Although the function ex is called the exponential function, in fact any function of the form ax is exponential.

y

x

y = 1x = 1

y = 0.5x

y = 1.5x

y = 2x

y = ex

y = 3xUse your graphical calculator to plotthe following:

1. y = 1.5x

2. y = 2x

3. y = ex

4. y = 3x

5. y = 1x

6. y = 0.5x

7. y = e-x

8. y = (-2)x

Discuss:Why are 6, 7 & 8 different to the rest?

The exponential function y = ex increases at an ever increasing rate. This is described as exponential growth.

By contrast, the graph of y = e-x, shown below, approaches the x axis ever more slowly as x increases: this is called exponential decay.


y

x

y = e-x

This works because:

y = e-x = (1/e)x.

1/e = 1/2.718281828 = 0.3678794412 < 1.

So, multiplying this by itself an increasing number of times will result in a smaller and smaller answer, similar to y = 0.5x.

You will meet ex and ln x again later in this course where you will learn to differentiate and integrate them. In this section we will be focusing on practical applications which require you to use the ln button on your calculator.



Example

The number, N, of insects in a colony is given by N = 2000e0.1t where t is the number of days after observations have begun.

(i) Sketch the graph of N against t.(ii) What is the population of the colony after 20 days?(iii) How long does it take the colony to reach a population of 10 000?


Solution


(i) Sketch the graph of N against t.

N

tO

N = 2000e0.1t

2000When t = 0, N = 2000e0 = 2000


Solution


(ii) What is the population of the colony after 20 days?

When t = 20, N = 2000e0.1×20 = 14 778.

The population is 14 778 insects.


Solution

The number, N, of insects in a colony is given by N = 2000e0.1t where t is the numberof days after observations have begun.

(iii) How long does it take the colony to reach a population of 10 000?

When N = 10 000, 10 000 = 2000e0.1t

5 = e0.1t.

Taking natural logarithms of both sides,

ln 5 = ln(e0.1t) = 0.1tand so t = 10ln5 = 16.09…

It takes just over 16 days for the population to reach 10 000.

Rememberln(ex) = x.


Example

The radioactive mass, M grams of a lump of material is given by M = 25e-0.0012t where t is the time in seconds since the first observation.

(i) Sketch the graph of M against t.(ii) What is the initial size of the mass?(iii) What is the mass after 1 hour?(iv) The half-life of a radioactive substance is the time it takes to decay to half of its

mass. What is the half-life of this material?


Solution


(i) Sketch the graph of M against t.

M

tO

25


Solution


(ii) What is the initial size of the mass?

When t = 0, M = 25e0

= 25.

The initial mass is 25g.


Solution


(iii) What is the mass after 1 hour?

After 1 hour, t = 3600. M = 25e-0.0012×3600.

The mass after 1 hour is 0.33g (to 2 decimal places).


Solution


(iv) The half-life of a radioactive substance is the time it takes to decay to half of its mass. What is the half-life of this material?

The initial mass is 25g, so after one half-life,

M = ½ × 25 = 12.5g.

At this point the value of t is given by 12.5 = 25e-0.0012t.

Dividing both sides by 25 gives 0.5 = e-0.0012t.

Taking logarithms of both sides: ln 0.5 = ln e-0.0012t

= -0.0012t.

t = = 577.6 (to 1 decimal place).

The half-life is 577.6 seconds.

ln 0.5-0.0012


Example

Make p the subject of ln(p) – ln(1 – p) = t.

Solution


p

1 - pln = t

Using log a – log b = log ab

Writing both sides as powers of e gives

ep

1 - pln = et

Remember elnx = x

p1 - p

= et

p = et(1 – p)

p = et – pet

p + pet = et

p(1 + et) = et

p =et

1 + et

Example

Make p the subject of ln(p) – ln(1 – p) = t.


Example

Solve the equation e1-3x = 5.


Example

Solve the equation e1-3x = 5.

Solution

Take natural logarithms of both sides: ln(e1-3x) = ln 5

Using the laws of logarithms: (1 – 3x)ln e = ln 5

Since ln e = 1: 1 – 3x = ln 5

x =1 – ln 53

= -0.203


Example

Solve the equation ln(2x + 1) = 3


Example

Solve the equation ln(2x + 1) = 3

Solution

ln(2x + 1) = 3 2x + 1 = e3

x =

e3 – 12

= 9.54

The exponential function e A reminder from C2. Use your calculator (if you need to) to work out the...

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