Exponential and Log functions

After completing this chapter you should be able to:

• Sketch simple transformations of the graph y = ex

• Sketch simple transformations of the graph y = ln x• Solve equations involving ex and ln x• Know what is meant by the terms exponential

growth and decay• Solve real life examples of exponential growth and

decay

Exponential functions are of the form y = ax

Why do all graphs of these functions pass through (0,1)?

Go to graph ( or a graphical calculator) and ploty = 1x

y = 2x

y = 3x

y = 4x

What do you notice ?What happens between 2 and 3?

between 2 and 3 there is a number such that the gradient function would be the same as the

function.

this number is represented by the letter ‘e’.

e ≈ 2.718

y = ex is often referred to as the exponential function

if y then

click here for more about e

population growth in real life is modelled by the graph of y = ex

All exponential graphs follow a similar pattern.

y = ex

if you were to draw up a table of values for this function you could see how rapidly it grows.

also note:

as x →∞ then ex→∞when x = 0, e0 = 1 (0,1) lies on the curveas x →-∞ then ex→ 0 (it approaches but never reaches the x axis)

X -2 -1 0 1 2 3 4 5y 0.14 0.37 1 2.1 7.4 20 55 148

How about the graph of y = e-x

this is a reflection of y = ex in the y-axis

this graph is often referred to as exponential decay.it is used to model examples from real life such as the fall in value of a car as well as the decay in radioactive isotopes.

Sketching exponential graphs

1. calculate some values to get an idea of the shape2. mark points where the axes are crossed

for example y = 10e-x

a table of values would give X -3 0 3

y 201 10 0.5

(0,10) y =10e-x

0

And the graph would look like this

Solving Problems involving exponential decay and growth

The price of a used car can be represented by the formula

where P is the price in £’s and t is the age in years from new.

Calculate:a) the new priceb) the value at 5 years old What does the model suggest about the eventual value of the car? Use this to sketch the graph of P against t

I will pause here while you work it out

you should have got:

a) when t = 0 then [remember e0 = 1] P = 16000 x 1 Price when new is £16000.

b) when t = 5 P = £9704.49 Price after 5 years is £9704.49c) as t →∞ → 0 therefore P → 0the eventual value is £0 16000

Exercise 3A page 35

To go further we now need to look at the inverse of ex.

The inverse function of ex is logex or lnx

With this information we can plough on and solve equations.

Solve the equation ex = 7Using the inverse operation gives

x = ln7

Now solve lnx = 25Taking the inverse gives

x = e25

A bit harder now:

Solve e2x+3 = 9Taking inverses give 2x + 3 = ln9

2x = ln9 – 3 x = ½ln9 - x = ln3 -

Remember½ln9 = ln9½ =ln3

And now solve2lnx + 1 = 5

Rearrange to get 2lnx = 4 lnx = 2 x = e2

We also need to be able to plot the graph of lnx and simple transformations of it

The important facts about the graph y = lnx are:

• as x →0 then y →∞

• when x = 1, y = 0 (0,1) lies on the curve

• lnx does not exist for negative numbers

• as x →∞ then y → ∞ (slowly)

Sketch the graph of y = ln(3 – x)

Thinking this through:when x → 3 , y → ∞when x = 2 y = ln(3 – 2) = ln1 = 0As x → -∞ , y → ∞ (slowly)

So the graph will look like this

How about the graph of y = 3 + ln(2x) ?

Thinking it through again: when x → 0 , y → ∞when x = ½, y = 3 + ln1 = 3As x → ∞ , y → ∞ (slowly)

So the graph will look like this

Now try exercise 3B page 39