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Prior Knowledge Check

1) Given that 𝑥 = 3 and 𝑦 = −1, evaluate these expressions without a calculator:

a) 5𝑥 b) 3𝑦 c) 22𝑥−1

d) 71−𝑦 e) 11𝑥+3𝑦

2) Simplify each expression by writing it as a single power:

a) 68 ÷ 62 b) 𝑦3 × 𝑦9 2

c) 25×29

28d) 𝑥8

3) Plot the following on a scatter graph and draw a line of best fit.

Find the gradient of your line of best fit, giving answers to 1dp.

𝒙 1.2 2.1 3.5 4 5.8

𝒚 5.8 7.4 9.4 10.3 12.8

1251

3 32

49 1

66 𝑦21

26 𝑥4

𝑦 = 1.5𝑥 + 4.1

Exponentials and LogarithmsGraphs 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

14A

xy a 0a where

84211/21/4

1/8y

3210-1-2-3x

Remember:32

3

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

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…

14A


Here are a few more examples of graphs where

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

14A

Imagine you have £100 in a bank account

Imagine your interest rate for the year is 100%

You will receive 100% interest in one lump at the end of the year, so you will now have £200 in the bank

However, you are offered a possible alternative way of being paid

Your bank manager says, ‘If you like, you can have your 100% interest split into two 50% payments, one made halfway through the year, and one made at the end’

How much money will you have at the end of the year, doing it this way (and what would be the quickest calculation to work that out?)

£100 x 1.52

= £225

Investigate further. What would happen if you split the interest into 4, or 10, or 100 smaller bits etc…

Exponentials and Logarithms

14B

£100e£100 x (1 + 1/n)n100/nn£100

£271.81£100 x 1.0001100000.01%10,000£100

£271.69£100 x 1.00110000.1%1,000£100

£270.48£100 x 1.011001%100£100

£269.16£100 x 1.02502%50£100

£265.33£100 x 1.05205%20£100

£259.37£100 x 1.11010%10£100

£256.58£100 x 1.125812.5%8£100

£244.14£100 x 1.25425%4£100

£225£100 x 1.5250%2£100

£200£100 x 2100%1£100

Total (2dp)SumInterest Each

PaymentPayments

Start Amount

11

n

en

The larger the value of n, the better the accuracy of e…

The value of e is irrational, like π…

It also has another interesting property…

(2.718281828459…)


14B

Gradient Functions

You have already learnt about differentiation, that differentiating a graph function gives the gradient function…

We can plot the gradient function on the graph itself…

y = x2

So dy/dx = 2x

y = x2

y = 2x

The Gradient at this point…

… is this value here!


14B

Gradient Functions



y = x3

So dy/dx = 3x2

y = x3y = 3x2

The Gradient at this point…

… is this value here!

And the Gradient is the

same here!


14B

Gradient Functions



y = 2x

dy/dx = 2xln2

You will learn where this comes from in

Year 13!

y = 2x

y = 2xln2


14B

Gradient Functions



y = 3x

dy/dx = 3xln3

You will learn where this comes from in

Year 13!

y = 3x

y = 3xln3


14B

y = 3xy = 3xln3y = 2x y = 2xln2

What has happened from the first graph to the second?

The lines have crossed…

Therefore the must be a value between 2 and 3 where the lines are equal…


14B

Gradient Functions

If we plot a graph of ex, its gradient function is the same graph!

This leads to an interesting conclusion…

If y = ex

Then dy/dx = ex as well!

y = ex

y = ex


14B


14B

If: 𝑓 𝑥 = 𝑒𝑥

Then: 𝑓′ 𝑥 = 𝑒𝑥

If: 𝑓 𝑥 = 𝑒𝑘𝑥

Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥

You should learn these two results – in Year 13 you will see where they come from!


14B

If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥

You need to be able to differentiate 𝒆𝒙 using the results above

Differentiate the following with respect to x:

a) 𝑦 = 𝑒2𝑥

b) 𝑦 = 𝑒−1

2𝑥

c) 𝑦 = 3𝑒2𝑥

You can do all of these using the pattern at the top right of the screen!

𝑦 = 𝑒2𝑥

𝑑𝑦

𝑑𝑥= 2𝑒2𝑥

𝑦 = 𝑒−12𝑥

𝑑𝑦

𝑑𝑥= −

1

2𝑒−

12𝑥

𝑦 = 3𝑒2𝑥

𝑑𝑦

𝑑𝑥= 6𝑒2𝑥

Follow the pattern!!

Follow the pattern!

Follow the pattern!!!

You need to be able to sketch transformations of the graph

𝒚 = 𝒆𝒙

y = ex

y = 2ex

y = ex

(0,1)f(x)

2f(x)

y = 2ex

(0,2)

(For the same set of inputs (x), the

outputs (y) double)


14B


y = ex

(0,1)f(x)

f(x) + 2

y = ex + 2

(0,3)


outputs (y) increase by 2)


𝒚 = 𝒆𝒙

y = ex

y = ex + 2

14B



y = ex

(0,1)f(x)

-f(x)y = -ex

(0,-1)


outputs (y) ‘swap signs’


𝒚 = 𝒆𝒙

y = ex

y = -ex

14B



y = ex

(0,1)f(x)

f(2x)

y = e2x

(The same set of outputs (y) for half

the inputs (x))


𝒚 = 𝒆𝒙

y = ex

y = e2x

14B



y = ex

(0,1)f(x)

f(x + 1)

y = ex + 1

(The same set of outputs (y) for inputs

(x) one less than before…)

(0,e)

We can work out the y-intercept by substituting in x = 0

This gives us e1 = e


𝒚 = 𝒆𝒙

y = ex

y = ex + 1

14B



y = ex

The graph of e-x, but with y values 10

times bigger…

y = e-x

y = 10e-x

(0, 1)

(0, 10)


𝒚 = 𝒆𝒙

y = ex

y = 10e-x

14B



y = ex

The graph of e0.5x, but with y values 4 times bigger with 3

added on at the end…

(0, 1)

(0, 7)

y = e0.5x

y = 4e0.5x

y = 3 + 4e0.5x

(0, 4)


𝒚 = 𝒆𝒙

y = ex

y = 3 + 4e0.5x

14B




You can use 𝒆𝒙 to model situations such as population growth, where the rate of change depends on the current amount

The density of a pesticide in a section of field, 𝑃 𝑚𝑔/𝑚2, can be modelled by the

equation:

𝑃 = 160𝑒−0.006𝑡

In this case, 𝑡 is the time in days since the pesticide was first applied.

a) Estimate the density of the pesticide after 15 days

b) Interpret the meaning of the 160 in this model

14C

𝑃 = 160𝑒−0.006𝑡

𝑃 = 160𝑒−0.006(15)

𝑃 = 146.2

Sub in 𝑡 = 15

Calculate

𝑃 = 160𝑒−0.006𝑡

𝑃 = 160𝑒−0.006(0)

𝑃 = 160

Sub in 𝑡 = 0

Calculate

When 𝑡 = 0, we get 160 as the answer

Using 𝑡 = 0 implies that no time has passed, so therefore 160 must be the original amount of

pesticide sprayed in the area

146.2𝑚𝑔



You can use 𝒆𝒙 to model situations such as population growth, where the rate of change depends on the current amount

The density of a pesticide in a section of field, 𝑃 𝑚𝑔/𝑚2, can be modelled by the

equation:

𝑃 = 160𝑒−0.006𝑡

c) Find 𝑑𝑃

𝑑𝑡

d) Interpret the significance of the sign of your answer to part c

e) Sketch the graph of 𝑃 against 𝑡.

14C


𝑃 = 160𝑒−0.006𝑡

𝑑𝑃

𝑑𝑡= −0.96𝑒−0.006𝑡

The sign is negative

This means that the gradient function (in this case) is downward sloping

In context, this means that the level of decay of the pesticide is decreasing

160

The original graph is a negative exponential shape!

Exponentials and LogarithmsWriting expressions as Logarithms

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

1) Write 25 = 32 as a logarithm

14D

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


means that

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

14D

loga n x xa n


means that

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

14D

loga n x xa n

Exponentials and LogarithmsLaws of logarithms

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

14E

log log loga a axy x y

log log loga a a

xx y

y

log ( ) logk

a ax k x

1log loga a x

x

(The Multiplication law)

(The Division law)

(The Power law)

Proof of the first rule:

Suppose that;

loga x b loga y cand

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’


Write each of these as a single logarithm:


log log loga a a

xx y

y

log ( ) logk

a ax k x

1log loga a x

x

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 3

5 5log 3 log 2

5 5log 9 log 8

5log (9 8)

5log 72

14E


Write each of these as a single logarithm:


log log loga a a

xx y

y

log ( ) logk

a ax k x

1log loga a x

x

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

14E


Write in terms of logax, logay and logaz


log log loga a a

xx y

y

log ( ) logk

a ax k x

1log loga a x

x

1) 2 3log ( )a x yz

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

2log log 3loga a ax y z

2)3

loga

x

y

3log log ( )a ax y

log 3loga ax y

14E


Write in terms of logax, logay and logaz


log log loga a a

xx y

y

log ( ) logk

a ax k x

1log loga a x

x

3) loga

x 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

14E


Solve the equation:

2𝑙𝑜𝑔2𝑥 = 8

log log loga a axy x y log log loga a a

xx y

y

log ( ) logk

a ax k x

14E

2𝑙𝑜𝑔2𝑥 = 8

𝑙𝑜𝑔2𝑥2 = 8

28 = 𝑥2

256 = 𝑥2

±16 = 𝑥

Use the power law

Write without the logarithm

Work out left side

Square root…

Check that your answers work…

2𝑙𝑜𝑔2(16)

= 8

2𝑙𝑜𝑔2(−16)

= 𝐸𝑟𝑟𝑜𝑟

You cannot calculate the logarithm of a negative value

In terms of graphs, the answer would be where the graph 𝑦 = 𝑎𝑥

is below the x-axis (ie 𝑦 < 0)


Solve the equation:

𝑙𝑜𝑔104 + 2𝑙𝑜𝑔10𝑥 = 2


xx y

y

log ( ) logk

a ax k x

14E

𝑙𝑜𝑔104 + 2𝑙𝑜𝑔10𝑥 = 2

𝑙𝑜𝑔104 + 𝑙𝑜𝑔10𝑥2 = 2

𝑙𝑜𝑔104𝑥2 = 2

102 = 4𝑥2

100 = 4𝑥2

25 = 𝑥2

5 = 𝑥

Use the power law

Use the addition law

Rewrite without the logarithm

Work out left side

Divide by 4

Square root

Note that the answer cannot be -5 (as in the previous example…


Solve the equation:

𝑙𝑜𝑔3 𝑥 + 11 − 𝑙𝑜𝑔3 𝑥 − 5 = 2

14E


xx y

y

log ( ) logk

a ax k x

𝑙𝑜𝑔3 𝑥 + 11 − 𝑙𝑜𝑔3 𝑥 − 5 = 2

𝑙𝑜𝑔3𝑥 + 11

𝑥 − 5= 2

32 =𝑥 + 11

𝑥 − 5

9 =𝑥 + 11

𝑥 − 5

9𝑥 − 45 = 𝑥 + 11

8𝑥 = 56

𝑥 = 7

Use the subtraction law

Rewrite without using the logarithm

Calculate left side

Multiply by (𝑥 − 5)

Rearrange

Divide by 8

Exponentials and LogarithmsSolving Equations using Logarithms

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

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 by default.

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..

14F

(3sf)

Calculate


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’

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

( 1)log7 ( 2)log3x x

log7 log7 log3 2log3x x

log7 log3 2log3 log7x x

(log7 log3) 2log3 log7x

2log3 log7

(log7 log3)x

0.297x

Bring the powers down

Multiply out the brackets

Rearrange to get ‘x’s together

Factorise to isolate the x term

Divide by (log7-log3)

Be careful when typing it all in!

(3dp)

14F


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’

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 5x

y2 = 52x

Factorise

You have 2 possible answers

‘Take logs’

Bring the power down

Divide by log5

Make sure it is accurate…

(2dp)14F


The graph of 𝒚 = 𝒍𝒏𝒙 is a reflection of the graph 𝒚 = 𝒆𝒙 in

the line 𝒚 = 𝒙.

The notation 𝑙𝑛 means the logarithm to the base of 𝑒, and is also on your

calculator.

𝑙𝑜𝑔𝑒𝑥 = 𝑙𝑛𝑥

We can compare the graphs of 𝑦 = 𝑒𝑥 and 𝑦 = 𝑙𝑛𝑥

14G

y = ex

y = lnx

y = x

(0,1)

(1,0)




You should use 𝑙𝑛 when solving exponential equations involving 𝑒.

Solve the equation 𝑒𝑥 = 5

14G

𝑒𝑥 = 5

𝑙𝑜𝑔𝑒𝑒𝑥 = 𝑙𝑜𝑔𝑒5


𝑥 = 𝑙𝑜𝑔𝑒5

𝑥𝑙𝑜𝑔𝑒𝑒 = 𝑙𝑜𝑔𝑒5

Take natural logs of both sides

Use the power law

𝑙𝑜𝑔𝑒𝑒 = 1

It is fine to leave your answers as exact values

𝑒𝑥 = 5

𝑙𝑛𝑒𝑥 = 𝑙𝑛5

𝑥 = 𝑙𝑛5

Take natural logs of both sides

Use the power law

You should use notation like the above when dealing with natural logarithms





Solve the equation 𝑙𝑛𝑥 = 3

14G

𝑙𝑛𝑥 = 3


Think about what ln represents

𝑙𝑜𝑔𝑒𝑥 = 3

𝑒3 = 𝑥

Rewrite without using a logarithm

𝑙𝑛𝑥 = 3Learn this pattern so you can use it !

𝑒3 = 𝑥





Solve the equation 𝑒2𝑥+3 = 7

14G


𝑒2𝑥+3 = 7

2𝑥 + 3 = 𝑙𝑛7

2𝑥 = 𝑙𝑛7 − 3

𝑥 =1

2𝑙𝑛7 −

3

2

Take natural logs

Divide by 2

Subtract 3





Solve the equation 2𝑙𝑛𝑥 + 1 = 5

14G


2𝑙𝑛𝑥 + 1 = 5Subtract 1

Write without a logarithm

Divide by 22𝑙𝑛𝑥 = 4

𝑙𝑛𝑥 = 2

𝑒2 = 𝑥





Solve the equation 𝑒2𝑥 + 5𝑒𝑥 = 14

14G


𝑒2𝑥 + 5𝑒𝑥 = 14

𝑒2𝑥 + 5𝑒𝑥 − 14 = 0

𝑒𝑥 + 7 𝑒𝑥 − 2 = 0

𝑒𝑥 = −7 or 𝑒𝑥 = 2

Subtract 14

𝑒𝑥 = 2

No solutions if negative

𝑥 = 𝑙𝑛2

Factorise (or write using y to help)

2 possible solutions for 𝑒𝑥

Take natural logs


Logarithms can be used to manage and explore non-linear trends in data

You are familiar with problems involving straight-line graphs

You can use logarithms to turn an exponential relationship into a linear

one

In short, using a logarithmic scale can turn a curved line into a straight line!

14H



In short, using a logarithmic scale can turn a curved line into a straight line!

14H

𝒙

𝒚 = 𝒙𝟐𝒚 𝒍𝒐𝒈𝒚 = 𝟐𝒍𝒐𝒈𝒙𝒍𝒐𝒈𝒚

𝒍𝒐𝒈𝒙

These are the same equation, one with logs having been taken

If it is plotted on logarithmic axes, it is a straight line



This can be shown by starting with an exponential relationship…

14H

𝑦 = 𝑎𝑥𝑛

𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎𝑥𝑛

𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑙𝑜𝑔𝑥𝑛

𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥

𝑦 = 𝑚𝑥 + 𝑐

Take logs of both sides

Separate using the addition law

Use the power law

We can compare this with the straight line

form…The ‘𝑦’ term is variable

We then have a constant multiplied by a variable

(𝑛 would be the constant)

We then have a constant

𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥



The data shows the rank (by size) and population of some UK cities.

The relationship between 𝑃 and 𝑅 can be modelled by the formula:

𝑃 = 𝑎𝑅𝑛

Where 𝑎 and 𝑛 are constants.

14H


City Birmingham Leeds Glasgow Sheffield Bradford

Rank, 𝑹 2 3 4 5 6

Population, 𝑷

1,000,000 730,000 620,000 530,000 480,000

A scatter graph of this data would look like this!

𝑹

𝑷





𝑃 = 𝑎𝑅𝑛


a) Draw a table giving values of 𝑙𝑜𝑔𝑅and 𝑙𝑜𝑔𝑃 to 2 decimal places

b) Plot a graph of 𝑙𝑜𝑔𝑅 against 𝑙𝑜𝑔𝑃 using the values from your table, and draw a

line of best fit

14H



Rank, 𝑹 2 3 4 5 6

Population, 𝑷

1,000,000 730,000 620,000 530,000 480,000


𝒍𝒐𝒈𝑹

𝒍𝒐𝒈𝑷

0.30 0.48 0.60 0.70 0.78

6 5.86 5.79 5.72 5.68

𝒍𝒐𝒈𝑹

𝒍𝒐𝒈𝑷



14H


𝑹

𝒍𝒐𝒈𝑷

𝒍𝒐𝒈𝑹

𝑷





𝑃 = 𝑎𝑅𝑛


b) Plot a graph of 𝑙𝑜𝑔𝑅 against 𝑙𝑜𝑔𝑃 using the values from your table, and draw a

line of best fit

c) Use your graph to estimate the values of 𝑎 and 𝑛 to two significant figures

First you need to rearrange the original relationship into the

𝑦 = 𝑚𝑥 + 𝑐 form.

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𝒍𝒐𝒈𝑹

𝒍𝒐𝒈𝑷

0.30 0.48 0.60 0.70 0.78

6 5.86 5.79 5.72 5.68

𝒍𝒐𝒈𝑷

𝒍𝒐𝒈𝑹





𝑃 = 𝑎𝑅𝑛



First you need to rearrange the original relationship into the

𝑦 = 𝑚𝑥 + 𝑐 form.

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𝑃 = 𝑎𝑅𝑛

𝑙𝑜𝑔𝑃 = 𝑙𝑜𝑔𝑎𝑅𝑛

𝑙𝑜𝑔𝑃 = 𝑙𝑜𝑔𝑎 + 𝑙𝑜𝑔𝑅𝑛

𝑙𝑜𝑔𝑃 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑅

Take logs

Addition law

Power law

So 𝑙𝑜𝑔𝑎 is the constant from the linear graph

And 𝑛 is the gradient from the linear graph





𝑃 = 𝑎𝑅𝑛



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So 𝑙𝑜𝑔𝑎 is the constant from the linear graph

And 𝑛 is the gradient from the linear graph

𝒍𝒐𝒈𝑷

𝒍𝒐𝒈𝑹

𝒍𝒐𝒈𝒂 = 𝟔. 𝟐

𝒏 = −𝟎. 𝟔𝟕

𝒂 = 𝟏, 𝟔𝟎𝟎, 𝟎𝟎𝟎

𝑃 = 𝑎𝑅𝑛 𝑃 = 1600000𝑅−0.67

𝟏𝟎𝟔.𝟐 = 𝒂



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𝑃 = 1600000𝑅−0.67

𝑹

𝑷

𝑹

𝑷

Original data Calculated graph



There is a second situation that you may need to deal with, which is slightly

different…

𝑦 = 𝑎𝑏𝑥

In this case, the variable x is in the power, rather than being raised to a

power…

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𝑦 = 𝑎𝑏𝑥

𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎𝑏𝑥

𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑙𝑜𝑔𝑏𝑥

𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑥𝑙𝑜𝑔𝑏

𝑦 = 𝑚𝑥 + 𝑐

Take logs of both sides

Separate using the addition law

Use the power law

We can compare this with the straight line

form…The ‘𝑦’ term is variable

We then have a constant multiplied

by a variable(𝑏 would be the

constant)

We then have a constant

𝑦 = 𝑎𝑏𝑥 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑥𝑙𝑜𝑔𝑏



Note that the axes we use for these will be different to the previous example…

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𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥 𝑦 = 𝑎𝑏𝑥 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑥𝑙𝑜𝑔𝑏

𝑦 = 𝑎𝑥𝑛

𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥

𝑦 = 𝑎𝑏𝑥


In this example, to get a linear relationship we plot

𝑙𝑜𝑔𝑦 against 𝑙𝑜𝑔𝑥

In this example, to get a linear relationship we plot

𝑙𝑜𝑔𝑦 against 𝑥

Ensure you check what form the equation is, as this will affect what you plot on the x-axis…



The graph shown represents the growth of a population of bacteria, 𝑃 over a period of 𝑡 hours. The graph has a

gradient of 0.6 and meets the vertical axis at (0,2) as shown.

A scientist suggests that this growth can be modelled by the equation 𝑃 = 𝑎𝑏𝑡,

where a and b are constants to be found.

a) Write down an equation for the line

b) Using your answer to part a or otherwise, find the values of 𝑎 and 𝑏, giving them to 3sf where necessary

c) Interpret the meaning of the constant 𝑎 in this model

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𝑙𝑜𝑔𝑃

𝑡

𝟐

𝑂

Use the form above as a starting point…


𝑙𝑜𝑔𝑃 = 𝑙𝑜𝑔𝑎 + 𝑡𝑙𝑜𝑔𝑏

𝑙𝑜𝑔𝑃 = 2 + 0.6𝑡

Use the variables in the question

The y-intercept is 𝑙𝑜𝑔𝑎 and the gradient is 𝑙𝑜𝑔𝑏

𝑙𝑜𝑔𝑃 = 2 + 0.6𝑡










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𝑙𝑜𝑔𝑃

𝑡

𝟐

𝑂

𝑙𝑜𝑔𝑃 = 2 + 0.6𝑡

𝑃 = 102+0.6𝑡

𝑙𝑜𝑔𝑃 = 2 + 0.6𝑡 𝑃 = 102 × 100.6𝑡

𝑃 = 102 × 100.6 𝑡

𝑃 = 100 × 3.98𝑡

𝑎 = 100 𝑏 = 3.98

Write without the logarithm

Separate using index laws

Rewrite the right part as a bracket

Calculate each power

You can compare this with the form suggested earlier…

𝑃 = 100 × 3.98𝑡










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𝑙𝑜𝑔𝑃

𝑡

𝟐

𝑂

𝑙𝑜𝑔𝑃 = 2 + 0.6𝑡

𝑃 = 100 × 3.98𝑡

𝑃 = 100 × 3.98𝑡

𝑃 = 100

Let 𝑡 = 0

Calculate

𝑃 = 100 × 3.980

So in this case, 100 gives the initial size of the bacteria population

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