The Exponential & Logarithmic Functions Exponential Growth & Decay Worked Example: Joan puts...

Higher Maths: Unit 3.3

The Exponential & Logarithmic Functions

Higher Maths: Unit 3.3Higher Maths: Unit 3.3Higher Maths: Unit 3.3Higher Maths: Unit 3.3Higher Maths: Unit 3.3Higher Maths: Unit 3.3Higher Maths: Unit 3.3Higher Maths: Unit 3.3Higher Maths: Unit 3.3

Exponential Growth & Decay

Worked Example:Joan puts £2500 into a savings account earning 13% interest per annum. How much money will she have if she leaves it there for 15 years?

Let £A(n) be the amount in her account after n years, then:

(0) 2500A 1(1) 2500 0.13 2500 2500 1.13A

2(2) (1) 0.13 (1) 2500 1.13A A A 3(3) (2) 0.13 (2) 2500 1.13A A A

( ) ( 1) 0.13 ( 1) 2500 1.13A n A n A n

nnnnnnn15(15) 2500 1.13 15635.68A

Example 1:The population of an urban district is decreasing at the rate of 2% per year.

(a) Taking P0 as the initial population, find a formula for the population, Pn, after n years.

(b) How long will it take for the population to drop from 900 000 to 800 000?

0 0 0(1) 0.02 0.98P P P P

1 1 1(2) 0.02 0.98P P P P 20 00.98 0.98 0.98P P

?

2 2 2(3) 0.02 0.98P P P P 2 30 00.98 0.98 0.98P P

Suggests that

00.98nnP P

(a)

Using our formula

00.98nnP P

And setting up the graphing calculator

In the fifth year the population drops below 800 000

Example 2:The rabbit population on an island increases by 15% each year. How many years will it take for the population to at least double?

0 0 0(1) 0.15 1.15P P P P

0PLet be the initial population

21 1 1 0(2) 0.15 1.15 1.15P P P P P

32 2 2 0(3) 0.15 1.15 1.15P P P P P

0( ) 1.15nP n P

Set 0 1P After 4 years the

population doubles

A Special Exponential Function – the “Number” e

The letter e represents the value 2.718….. (a never ending decimal). This number occurs often in nature

f(x) = 2.718..x = ex is called the exponential function to the base e.

Although we now think of logarithms as the exponents to which one must raise the base to get the required number, this is a modern way of thinking. In 1624 e almost made it into the mathematical literature, but not quite. In that year Briggs gave a numerical approximation to the base 10 logarithm of e but did not mention e itself in his work. Certainly by 1661 Huygens understood the relation between the rectangular hyperbola and the logarithm. He examined explicitly the relation between the area under the rectangular hyperbola yx = 1 and the logarithm. Of course, the number e is such that the area under the rectangular hyperbola from 1 to e is equal to 1. This is the property that makes e the base of natural logarithms, but this was not understood by mathematicians at this time, although they were slowly approaching such an understanding

Example 3:The mass of a fixed quantity of radioactive substance decays according to the formula m = 50e-0.02t, where m is the mass in grams and t is the time in years.

What is the mass after 12 years?

Linking the Exponential Function and the Logarithmic Function

y

x

( ) 2xf x

1( )f x (0,1)

(1,0)

In chapter 2.2 we found that the exponential function has an inverse function, called the logarithmic function.

log 1 0

log 1

log

a

a

xa

a

y a x y

2log x

The log function is the inverse of the exponential function, so it ‘undoes’ the exponential function:

f(x) = 2x

ask yourself:

1 2 21 = 2 so log22 = 1 “2 to what power gives 2?”

2 4 22 = 4 so log24 = “2 to what power gives 4?”

3 8 23 = 8 so log28 = “2 to what power gives 8?”

4 16 24 = 16 so log216 = “2 to what power gives 16?”

f(x) = log2x

23

4

Example 4:

(a)log381 = “…. to what power gives ….?”

(b)log42 = “…. to what power gives ….?”

1

27

(c)log3 = “…. to what power gives ….?”

4 3 81

4 2

-3 3

1

21

27

Rules of Logarithms

Rules for Logs:Rules for indices:

m n m na a a

m n m na a a

nm m na a

log log loga a axy x y

log log loga a a

xx y

y

log logpa ax p x

Example 5:

Simplify:

a) log102 + log10500 b) log363 – log37

10log (2 500)

10log 10003

Since310 1000

3

63log

7

3log 9

2

Since23 9

c) 2 2

1 1log 16 log 8

2 3

1

3

1

2

1

21

2

1

21

2 1

2

2 2log 16 log 8 1

3

1

2

1

3

1

31

3

1

3 3

2 2log 16 log 8 2 2log 4 log 2 2 1 1

Since2log 4 2

Since2log 2 1

Using your Calculator

You have 2 logarithm buttons on your calculator:

which stands for log10 and its inverse

which stands for loge and its inverse

log log

10x

ln

xe

ln

Try finding log10100on your calculator 2

Solving Exponential Equations

Solve 5 11x 51 = 5 and 52 = 25 so we can see that x lies between ……and…………..1 2

Taking logs of both sides and applying the rules

10 10log 5 log 11x xxxxxxxx 10 10log 5 log 11

10

10

log 111.489

log 5x

For the formula P(t) = 50e-2t:a) evaluate P(0)

b) for what value of t is P(t) = ½P(0)?

2 0(0) 50 50P e a)

b)1 1

(0) 50 252 2P

225 50 te21

2te

Could we have known this?

21ln ln

2te

21ln ln

2te

0.693 2 lnt e

0.693 2 1t 0.693

2t

0.346 t

The formula A = A0e-kt gives the amount of a radioactive substance after time t minutes. After 4 minutes 50g is reduced to 45g.(a) Find the value of k to two significant figures.

(b) How long does it take for the substance to reduce to half it original weight?

Example

(a) 4t (0) 50A (4) 45A

445 50 keTake logs of

both sides 4ln(45) ln 50 ke 4ln(45) ln 50 ln ke 4ln 45 ln 50 ln ke

4ln 45 ln 50 ln ke 45

ln 4 ln50

k e

0.1054 4k remember

ln 1e

0.1054

4k

0.0263 k

(b) How long does it take for the substance to reduce to half it original weight?

0.02631(0) (0)

2tA A e

0.02631

2te

0.02631ln ln

2te

0.693 0.0263 lnt e

0.693 0.0263 26.35t t

Experiment and Theory

When conducting an experiment scientists may analyse the data to find if a formula connecting the variables exists. Data from an experiment may result in a graph of the form shown in the diagram, indicating exponential growth. A graph such as this implies a formula of the type y = kxn

2 2 4 6 8 10

y

x

We can find this formula by using logarithms:

ny kxIf

Then log log ny kx

So log log ny kx log k log nx

log log ny kx log k logn xCompare this to Y mX c

log y Y m n logc k

So log log ny kx log k logn xIs the equation of a straight line

log y

log x

log y

log x

From ny kx

We see by taking logs that we can reduce this problem to a straight line problem where:

So log log ny kx log k logn x

Andlog y Y m n logc k

Worked Example:

The following data was collected during an experiment:

x 50.1 194.9 501.2 707.9

y 20.9 46.8 83.2 102.3

a) Show that y and x are related by the formula y = kxn

.

b) Find the values of k and n and state the formula that

connects x and y.

a) Taking logs of x and y gives:

logx

logy

1.69 2.28 2.70 2.841.32 1.67 1.92 2.00

Plotting these points

0.5

0.5

1

1

1.5

1.5

2

2

2.5

2.5

3

3

0.5

0.5

1

1

1.5

1.5

2

2

2.5

2.5

3

3

log10 Y

log10 X

We get a straight line and hence the formula connecting X and Y is of the form

y mx c

b)Since the points lie on a straight line, we can say that:

ny kxIf

Then log log ny kx

So log log ny kx log k log nx

log log ny kx log k logn xCompare this to Y mX c

By selecting points on the graph and substituting into this equation we get using

1.69,1.32 2.84,2.00

1.32 1.69m c 2.00 2.84m c

Subtract

0.68 1.15m0.68

1.15m

0.6 m

1.32 1.69 0.6 c

So

0.3 c

So we have 0.6 0.3Y X Compare this to

log log ny kx log k logn x

n and log k 0.6 0.3so

solving

0.310k

log k 0.3

1.99k

so 0.61.99ny kx y x

You can always check this on your graphics calculator