Exponentials and Logarithms This chapter is focused on functions which are exponential These...

Exponentials and Logarithms

• This chapter is focused on functions which are exponential

• These functions change at an increasing/decreasing rate

• Logarithms are used to solve problems involving exponential functions


Graphs of Exponential Functions

You need to be familiar with the function;

For example, y = 2x, y = 5x and so on…

1) Draw the graph of y = 2x

3A

xy a 0a where

84211/21/4

1/8y

3210-1-2-3x

Remember:

323

1

2

x

y

1

2

3

4

5

7

6

8

1 2 3-1-2-3

Any graph of will be the same basic shape

It always passes through (0,1) as anything to the power 0 is equal to 1

xy a



Here are a few more examples of graphs where

3A

xy a

0

5

10

15

20

25

30

-3 -2 -1 0 1 2 3

x

y

y = 3x

y = 2x

y = 1.5x

All pass through (0,1)

They never go below 0

Notice that either side of (0,1), the biggest/smallest

values switch

Above (0,1), y = 3x is the biggest

value, below (0,1), it is the

smallest…



Here are a few more examples of graphs where

3A

xy a

0

1

2

3

4

5

6

7

8

9

-3 -2 -1 0 1 2 3

x

y

y = 2x

y = (1/2)x

The graph y = (1/2)x is

a reflection of y = 2x

1

2

x

y

12x

y

2 xy


Writing expressions as Logarithms

‘a’ is known as the ‘base’ of the logarithm…

1) Write 25 = 32 as a logarithm …

3B

loga n x xa nmeans that

52 32

2log 32 5

Effectively, the 2 stays as the ‘first’

number…

The 32 and the 5 ‘switch positions’

2) Write as a logarithm:

a) 103 = 1000

b) 54 = 625

c) 210 = 1024

310 1000

10log 1000 3

45 625

5log 625 4

102 1024

2log 1024 10



3B


Find the value of:

a) 3log 81

What power do I raise 3 to, to get

81?

3log 81 4

b) 4log 0.25

What power do I raise 4 to, to get

0.25?

4log 0.25 1 0.25 is 1/4

Remember, 14 1

4



3B


Find the value of:

c) 0.5log 4

What power do I raise 0.5 to, to

get 4?

0.5log 4 2

d) 5log ( )a a

What power do I raise ‘a’ to, to get

a5?

5log ( ) 5a a 0.5 = 1/2

0.52 = 1/4

0.5-2 = 4


Calculating logarithms on a Calculator

On your calculator, you can calculate a logarithm.

Using the log button on the calculator automatically chooses base 10, ie) log20 will work out what power you must raise 10 to, to get 20

To work out log20, all you do is type log20 into the calculator!

log20 = 1.301029996…. 1.30 to 3sf

3C


Laws of logarithms

You do not need to know proofs of these rules, but you will need to learn and use them:

3D

log log loga a axy x y

log log loga a a

xx y

y

log ( ) logka ax k x

1log loga a xx

(The Multiplication law)

(The Division law)

(The Power law)

Proof of the first rule:

Suppose that;loga x b loga y can

d

ba x ca y

xy b ca a

xy b ca

loga xy b c ‘a must be raised to the power

(b+c) to get xy’


Laws of logarithms

Write each of these as a single logarithm:

3D


log log loga a a

xx y

y


1log loga a xx

1) 3 3log 6 log 7

3log (6 7)

3log 42

2) 2 2log 15 log 3

2log (15 3)

2log 5

3) 5 52log 3 3log 2

2 35 5log 3 log 2

5 5log 9 log 8

5log (9 8)

5log 72


Laws of logarithms

Write each of these as a single logarithm:

3D


log log loga a a

xx y

y


1log loga a xx

4) 10 10

1log 3 4log

2

4

10 10

1log 3 log

2

10 10

1log 3 log

16

10

1log 3

16

10log 48

Alternatively, using rule 4

10 10log 3 log 16

10log (3 16)

10log 48


Laws of logarithms

Write in terms of logax, logay and logaz

3D


log log loga a a

xx y

y


1log loga a xx

1) 2 3log ( )a x yz

2 3log ( ) log log ( )a a ax y z

2log log 3loga a ax y z

2) 3loga

x

y

3log log ( )a ax y

log 3loga ax y


Laws of logarithms

Write in terms of logax, logay and logaz

3D


log log loga a a

xx y

y


1log loga a xx

3) logax y

z

4)

log log loga a ax y z

1

2log log ( ) loga a ax y z

1log log log

2a a ax y z

4loga

x

a

4log log ( )a ax a

log 4loga ax a

log 4a x = 1


Solving Equations using Logarithms

Logarithms allow you to solve equations where ‘powers’ are involved.

You need to be able to solve these by ‘taking logs’ of each side of the equation.

All logarithms you use on the calculator will be in base 10.

3 20x

10 10log (3 ) log 20x

10 10log 3 log 20x

10

10

log 20

log 3x

1.3010...

0.4771...x

2.73x

‘Take logs’

You can bring the power

down…

Divide by log103

Make sure you use the exact

answers to avoid rounding errors..

3E

(3sf)



The steps are essentially the same when the power is an expression, such as ‘x – 2’, ‘2x + 4’ etc…

There is more rearranging to be done though, as well as factorising.

Overall, you are trying to get all the ‘x’s on one side and all the logs on the other…

1 27 3x x ‘Take logs’

3E

1 2log(7 ) log(3 )x x

( 1) log 7 ( 2) log3x x

log 7 log 7 log3 2log3x x

log 7 log3 2log3 log 7x x

(log 7 log3) 2log3 log 7x

2log3 log 7

(log 7 log3)x

0.297x

Bring the powers down

Multiply out the brackets

Rearrange to get ‘x’s together

Factorise to isolate the x

termDivide by (log7-log3)

Be careful when typing it

all in! (3dp)



You may also need to use a substitution method with even harder ones.

You will know to use this when you see a logarithm that has a similar shape to a quadratic equation..

Let y=5x

When you raise a number to a power, the answer cannot be negative…

25 7(5 ) 30 0x x Sub in ‘y = 5x’

3E

2 7 30 0y y

( 10)( 3) 0y y

10y 3y or

5 3x log5 log3x

log5 log3x log3

log5x

0.68x

y2 = 5x x 5xy2 = 52x

Factorise

You have 2 possible answers

‘Take logs’

Bring the power down

Divide by log5

Make sure it is accurate…

(2dp)


Changing the base

Your calculator will always give you answers for log10, unless you say otherwise.

You need to be able to change the base if your calculator cannot do this

You also need to be able to change the base to solve some logarithmic equations

3F

loga x m

ma x

log ( )mb a log ( )b x

logbm a logb x

m log

logb

b

x

a

loga x log

logb

b

x

a

Rewrite as an equation

‘Take logs’ to a different base

The power law – bring the m

down

Divide by logba

Sub in logax for m (from first

line)


Changing the base

3F

loga x log

logb

b

x

a

4log 9 10

10

log 9

log 4

4log 9 1.58 (2dp)

Special case

logab x log x

b

a

25log 10 10

5

log 2

43log 10 10

3

log 4


Changing the base

Find the value of log811 to 3.s.f

3F

loga x log

logb

b

x

a

8log 11 10

10

log 11

log 8

8log 11 1.15 (3sf)

Alternatively…8log 11

8 11x

10 10log (8 ) log 11x

10 10log 8 log 11x

10

10

log 11

log 8x

‘Take logs’

Power law

Divide by log108


Changing the base

Solve the equation:

log5x + 6logx5 = 5

3F

loga x log

logb

b

x

a

logab x log x

b

a

5log 6log 5 5xx

5log x5

6

log x

y 6

y5

2y 6 5y

5

2 5 6 0y y

( 2)( 3) 0y y

2 or 3y y

5log 2x

5log 3x

25x

125x

25 x

35 x

Use the ‘special case’ rule

Let log5x = y

Multiply by y

Rearrange like a quadratic

Factorise

Solve for y

Summary

• We have learnt what logarithms are

• We have learnt a number of rules which can be used to manipulate logarithms

• We have also seen how logarithms can help us solve equations with powers as unknowns

Exponentials and Logarithms This chapter is focused on functions which are exponential These...

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Transcript of Exponentials and Logarithms This chapter is focused on functions which are exponential These...