C3 - ln x and ex

Natural Logarithms and Exponentials

Natural Logs

We know that , except for n=-1 (i.e., we do not know how

to evaluate or ) This integral must have a value because if we

look at the graph of and consider the definite integral which

represents the area under the curve between and it is clear from

the graph that this must have a value, i.e.

In fact, the integral turns out to be a log function, which we call the natural log function. This has the base . On your calculator the natural log is

denoted by lnx.

The graph of lnx has the characteristic shape of all log functions:

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1 3

1

x

y

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We can transform the graph using translation, reflection and stretch eg

(i) y= ln(x+3) (ii) y = - lnx (iii) y=ln(2x)

The Exponential Function e x : We know that the inverse of is

. Similarly, the inverse of y = Inx is .

The graphs of inverse functions are the reflections of each other in the line y = x, so the graph of y = ex must look like this:

It also follows, since ff-1(x) = f-1f(x) = x, that

and LEARN

The graph can also be translated, reflected and stretched, eg(i) y = ex + 3 (ii) y = e-x (iii) y = e2x

Exponential growth and Decay

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1

1

y = ex

y = Inx

- 3 1 0.5

y = x

3 1 1

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Example of exponential growth.

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

i) Sketch he graph of N against tii) What is the population of the colony after 20 days?iii) How long does it take the colony to reach a population of 10000?

Solution i) When t=0, N=2000, and the graph is an x-stretch of

scale factor 10 of the graph y = ex. It therefore looks like this:

ii) When t=20, N=2000e =

iii) When N=10000, 10000= 2000e

5 = e

Taking logs of both sides (natural logs since ex is involved) In 5 = ln e0.1t

ln 5 = 0.1t

t =

It takes just over 16 days for the population to reach 10000.

Example of exponential decay

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2000

t

N

C3 - ln x and ex

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

i) Sketch the graph of M against tii) 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) When t=0, M = 25, and e-0.0012t is a reflection of et in the M axis with a

t-stretch of scale factor 833 . The graph must therefore look like

this:

ii) When initial mass =25g

iii) After 1 hour t = 3600.

(to 2 dp)

iv) After 1 half life,

In 0.5 = ln e-0.0012t

In (to 1 dp)

The half life is 557.6 seconds, just under 10 minutes Very highly radioactive.

Solving equations with e x and Inx

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C3 - ln x and ex

To solve an equation with ex, first combine all the e functions into a single e function on one side of the equation, then ‘take ln’ of both sides.

To solve an equation with ln x, first combine all the ln functions into a single ln function on one side of the equation, then ‘take e’ of both sides.

Examples Solve the equations (i) ex x e2x = 3 (ii) lnx – ln3 = lnx2

Solutions

(i) ex x e2x = 3ex+2x = 3e3x = 3

lne3x = ln33x = ln3

x = = 0.366(3sf)

(ii) lnx – ln3 = lnx2

lnx – ln3 –lnx2 = 0

ln [ ] = 0

= e0

= 1

x = 3x2

x - 3x2 = 0 x(1 – 3x) = 0

x = 0, x =

Example: Find an expression for p if

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C3 - ln x and ex

Solution

‘Take exp’ of both sides

Differentiating Inx and ex

Since ,

it follows that

The derivative of y = ex may be found by interchanging x and y and finding

from , ie

.

Therefore,

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LEARN

C3 - ln x and ex

So, is unchanged when differentiated or integrated. LEARN

Example 1: Find if y

Solution:

Example 2: Find if y=e

This is a function of a function use substitution + chain rule.

Solution

In general, LEARN

Example 3 Differentiate

Solution:

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Example 4 Differentiate

Solution

Example 5

Differentiate the following functions:

(a) y=2Inx (b) y=In(3x)

Solution

(a) (b) Let t = Int.

An alternative solution to (b) is In 3x =In3 + Inx

=

In general, LEARN

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Example 6

Differentiate the following functions:

a) y= In b) y= In

Solution

a) y= In In x

b) y=In

Let t = In t

Example 7: Differentiate y =

Solution: Use quotient rule.

Example 8

Differentiate

Solution: Use the product rule

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Integrals involving the exponential function

Since you know that

You can see that

LEARN

Also, if you use integration by substitution,

LEARN

Example: Find the following integrals;

a) b) c) d)

Solution a)

c)

d)

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Integrals involving Inx

N.B: note the mod x sign i.e always use the + value of x

Example1 : Evaluate

Solution:

Example 2: Find

Solution;

.

Example 3: Find

Solution:

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C3 - ln x and ex

Example 4: Find

Solution: Using integration by substitution,

In General,

LEARN

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