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Twitter: @Owen134866
www.mathsfreeresourcelibrary.com
Prior Knowledge Check
1) Given that ๐ฅ = 3 and ๐ฆ = โ1, evaluate these expressions without a calculator:
a) 5๐ฅ b) 3๐ฆ c) 22๐ฅโ1
d) 71โ๐ฆ e) 11๐ฅ+3๐ฆ
2) Simplify each expression by writing it as a single power:
a) 68 รท 62 b) ๐ฆ3 ร ๐ฆ9 2
c) 25ร29
28d) ๐ฅ8
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.
๐ 1.2 2.1 3.5 4 5.8
๐ 5.8 7.4 9.4 10.3 12.8
1251
3 32
49 1
66 ๐ฆ21
26 ๐ฅ4
๐ฆ = 1.5๐ฅ + 4.1
Exponentials and LogarithmsGraphs of Exponential Functions
You need to be familiar with the function;
For example, y = 2x, y = 5x and so onโฆ
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
Exponentials and LogarithmsGraphs of Exponential Functions
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 smallestโฆ
14A
Exponentials and LogarithmsGraphs of Exponential Functions
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 ยฃ100 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 ยฃ200 in the bank
However, you are offered a possible alternative way of being paid
Your bank manager says, โIf you like, you can have your 100% interest split into two 50% payments, one made halfway through the year, and one made at the endโ
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?)
ยฃ100 x 1.52
= ยฃ225
Investigate further. What would happen if you split the interest into 4, or 10, or 100 smaller bits etcโฆ
Exponentials and Logarithms
14B
ยฃ100eยฃ100 x (1 + 1/n)n100/nnยฃ100
ยฃ271.81ยฃ100 x 1.0001100000.01%10,000ยฃ100
ยฃ271.69ยฃ100 x 1.00110000.1%1,000ยฃ100
ยฃ270.48ยฃ100 x 1.011001%100ยฃ100
ยฃ269.16ยฃ100 x 1.02502%50ยฃ100
ยฃ265.33ยฃ100 x 1.05205%20ยฃ100
ยฃ259.37ยฃ100 x 1.11010%10ยฃ100
ยฃ256.58ยฃ100 x 1.125812.5%8ยฃ100
ยฃ244.14ยฃ100 x 1.25425%4ยฃ100
ยฃ225ยฃ100 x 1.5250%2ยฃ100
ยฃ200ยฃ100 x 2100%1ยฃ100
Total (2dp)SumInterest Each
PaymentPayments
Start Amount
11
n
en
The larger the value of n, the better the accuracy of eโฆ
The value of e is irrational, like ฯโฆ
It also has another interesting propertyโฆ
(2.718281828459โฆ)
Exponentials and Logarithms
14B
Gradient Functions
You have already learnt about differentiation, that differentiating a graph function gives the gradient functionโฆ
We can plot the gradient function on the graph itselfโฆ
y = x2
So dy/dx = 2x
y = x2
y = 2x
The Gradient at this pointโฆ
โฆ is this value here!
Exponentials and Logarithms
14B
Gradient Functions
You have already learnt about differentiation, that differentiating a graph function gives the gradient functionโฆ
We can plot the gradient function on the graph itselfโฆ
y = x3
So dy/dx = 3x2
y = x3y = 3x2
The Gradient at this pointโฆ
โฆ is this value here!
And the Gradient is the
same here!
Exponentials and Logarithms
14B
Gradient Functions
You have already learnt about differentiation, that differentiating a graph function gives the gradient functionโฆ
We can plot the gradient function on the graph itselfโฆ
y = 2x
dy/dx = 2xln2
You will learn where this comes from in
Year 13!
y = 2x
y = 2xln2
Exponentials and Logarithms
14B
Gradient Functions
You have already learnt about differentiation, that differentiating a graph function gives the gradient functionโฆ
We can plot the gradient function on the graph itselfโฆ
y = 3x
dy/dx = 3xln3
You will learn where this comes from in
Year 13!
y = 3x
y = 3xln3
Exponentials and Logarithms
14B
y = 3xy = 3xln3y = 2x y = 2xln2
What has happened from the first graph to the second?
The lines have crossedโฆ
Therefore the must be a value between 2 and 3 where the lines are equalโฆ
Exponentials and Logarithms
14B
Gradient Functions
If we plot a graph of ex, its gradient function is the same graph!
This leads to an interesting conclusionโฆ
If y = ex
Then dy/dx = ex as well!
y = ex
y = ex
Exponentials and Logarithms
14B
Exponentials and Logarithms
14B
If: ๐ ๐ฅ = ๐๐ฅ
Then: ๐โฒ ๐ฅ = ๐๐ฅ
If: ๐ ๐ฅ = ๐๐๐ฅ
Then: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
You should learn these two results โ in Year 13 you will see where they come from!
Exponentials and Logarithms
14B
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
You need to be able to differentiate ๐๐ using the results above
Differentiate the following with respect to x:
a) ๐ฆ = ๐2๐ฅ
b) ๐ฆ = ๐โ1
2๐ฅ
c) ๐ฆ = 3๐2๐ฅ
You can do all of these using the pattern at the top right of the screen!
๐ฆ = ๐2๐ฅ
๐๐ฆ
๐๐ฅ= 2๐2๐ฅ
๐ฆ = ๐โ12๐ฅ
๐๐ฆ
๐๐ฅ= โ
1
2๐โ
12๐ฅ
๐ฆ = 3๐2๐ฅ
๐๐ฆ
๐๐ฅ= 6๐2๐ฅ
Follow the pattern!!
Follow the pattern!
Follow the pattern!!!
You need to be able to sketch transformations of the graph
๐ = ๐๐
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)
Exponentials and Logarithms
14B
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
y = ex
(0,1)f(x)
f(x) + 2
y = ex + 2
(0,3)
(For the same set of inputs (x), the
outputs (y) increase by 2)
You need to be able to sketch transformations of the graph
๐ = ๐๐
y = ex
y = ex + 2
14B
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
Exponentials and Logarithms
y = ex
(0,1)f(x)
-f(x)y = -ex
(0,-1)
(For the same set of inputs (x), the
outputs (y) โswap signsโ
You need to be able to sketch transformations of the graph
๐ = ๐๐
y = ex
y = -ex
14B
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
Exponentials and Logarithms
y = ex
(0,1)f(x)
f(2x)
y = e2x
(The same set of outputs (y) for half
the inputs (x))
You need to be able to sketch transformations of the graph
๐ = ๐๐
y = ex
y = e2x
14B
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
Exponentials and Logarithms
y = ex
(0,1)f(x)
f(x + 1)
y = ex + 1
(The same set of outputs (y) for inputs
(x) one less than beforeโฆ)
(0,e)
We can work out the y-intercept by substituting in x = 0
This gives us e1 = e
You need to be able to sketch transformations of the graph
๐ = ๐๐
y = ex
y = ex + 1
14B
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
Exponentials and Logarithms
y = ex
The graph of e-x, but with y values 10
times biggerโฆ
y = e-x
y = 10e-x
(0, 1)
(0, 10)
You need to be able to sketch transformations of the graph
๐ = ๐๐
y = ex
y = 10e-x
14B
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
Exponentials and Logarithms
y = ex
The graph of e0.5x, but with y values 4 times bigger with 3
added on at the endโฆ
(0, 1)
(0, 7)
y = e0.5x
y = 4e0.5x
y = 3 + 4e0.5x
(0, 4)
You need to be able to sketch transformations of the graph
๐ = ๐๐
y = ex
y = 3 + 4e0.5x
14B
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
Exponentials and Logarithms
Exponentials and Logarithms
You can use ๐๐ 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, ๐ ๐๐/๐2, can be modelled by the
equation:
๐ = 160๐โ0.006๐ก
In this case, ๐ก 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
๐ = 160๐โ0.006๐ก
๐ = 160๐โ0.006(15)
๐ = 146.2
Sub in ๐ก = 15
Calculate
๐ = 160๐โ0.006๐ก
๐ = 160๐โ0.006(0)
๐ = 160
Sub in ๐ก = 0
Calculate
When ๐ก = 0, we get 160 as the answer
Using ๐ก = 0 implies that no time has passed, so therefore 160 must be the original amount of
pesticide sprayed in the area
146.2๐๐
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
Exponentials and Logarithms
You can use ๐๐ 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, ๐ ๐๐/๐2, can be modelled by the
equation:
๐ = 160๐โ0.006๐ก
c) Find ๐๐
๐๐ก
d) Interpret the significance of the sign of your answer to part c
e) Sketch the graph of ๐ against ๐ก.
14C
If: ๐ ๐ฅ = ๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐ฅ If: ๐ ๐ฅ = ๐๐๐ฅThen: ๐โฒ ๐ฅ = ๐๐๐๐ฅ
๐ = 160๐โ0.006๐ก
๐๐
๐๐ก= โ0.96๐โ0.006๐ก
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
โaโ is known as the โbaseโ of the logarithmโฆ
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 โfirstโ
numberโฆ
The 32 and the 5 โswitch positionsโ
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
Exponentials and LogarithmsWriting expressions as Logarithms
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
Exponentials and LogarithmsWriting expressions as Logarithms
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 โaโ 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
โa must be raised to the power (b+c) to get xyโ
Exponentials and LogarithmsLaws of logarithms
Write each of these as a single logarithm:
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
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
Exponentials and LogarithmsLaws of logarithms
Write each of these as a single logarithm:
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
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
Exponentials and LogarithmsLaws of logarithms
Write in terms of logax, logay and logaz
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
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
Exponentials and LogarithmsLaws of logarithms
Write in terms of logax, logay and logaz
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
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
Exponentials and LogarithmsLaws of logarithms
Solve the equation:
2๐๐๐2๐ฅ = 8
log log loga a axy x y log log loga a a
xx y
y
log ( ) logk
a ax k x
14E
2๐๐๐2๐ฅ = 8
๐๐๐2๐ฅ2 = 8
28 = ๐ฅ2
256 = ๐ฅ2
ยฑ16 = ๐ฅ
Use the power law
Write without the logarithm
Work out left side
Square rootโฆ
Check that your answers workโฆ
2๐๐๐2(16)
= 8
2๐๐๐2(โ16)
= ๐ธ๐๐๐๐
You cannot calculate the logarithm of a negative value
In terms of graphs, the answer would be where the graph ๐ฆ = ๐๐ฅ
is below the x-axis (ie ๐ฆ < 0)
Exponentials and LogarithmsLaws of logarithms
Solve the equation:
๐๐๐104 + 2๐๐๐10๐ฅ = 2
log log loga a axy x y log log loga a a
xx y
y
log ( ) logk
a ax k x
14E
๐๐๐104 + 2๐๐๐10๐ฅ = 2
๐๐๐104 + ๐๐๐10๐ฅ2 = 2
๐๐๐104๐ฅ2 = 2
102 = 4๐ฅ2
100 = 4๐ฅ2
25 = ๐ฅ2
5 = ๐ฅ
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 exampleโฆ
Exponentials and LogarithmsLaws of logarithms
Solve the equation:
๐๐๐3 ๐ฅ + 11 โ ๐๐๐3 ๐ฅ โ 5 = 2
14E
log log loga a axy x y log log loga a a
xx y
y
log ( ) logk
a ax k x
๐๐๐3 ๐ฅ + 11 โ ๐๐๐3 ๐ฅ โ 5 = 2
๐๐๐3๐ฅ + 11
๐ฅ โ 5= 2
32 =๐ฅ + 11
๐ฅ โ 5
9 =๐ฅ + 11
๐ฅ โ 5
9๐ฅ โ 45 = ๐ฅ + 11
8๐ฅ = 56
๐ฅ = 7
Use the subtraction law
Rewrite without using the logarithm
Calculate left side
Multiply by (๐ฅ โ 5)
Rearrange
Divide by 8
Exponentials and LogarithmsSolving Equations using Logarithms
Logarithms allow you to solve equations where โpowersโ are involved as unknowns.
You need to be able to solve these by โtaking logsโ 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
โTake logsโ
You can bring the power downโฆ
Divide by log103
Make sure you use the exact
answers to avoid rounding errors..
14F
(3sf)
Calculate
Exponentials and LogarithmsSolving Equations using Logarithms
The steps are essentially the same when the power is an expression, such as โx โ 2โ, โ2x + 4โ etcโฆ
There is more rearranging to be done though, as well as factorising.
Overall, you are trying to get all the โxโs on one side and all the logs on the otherโฆ
1 27 3x x โTake logsโ
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 โxโs together
Factorise to isolate the x term
Divide by (log7-log3)
Be careful when typing it all in!
(3dp)
14F
Exponentials and LogarithmsSolving Equations using Logarithms
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 negativeโฆ
25 7(5 ) 30 0x x Sub in โy = 5xโ
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
โTake logsโ
Bring the power down
Divide by log5
Make sure it is accurateโฆ
(2dp)14F
Exponentials and Logarithms
The graph of ๐ = ๐๐๐ is a reflection of the graph ๐ = ๐๐ in
the line ๐ = ๐.
The notation ๐๐ means the logarithm to the base of ๐, and is also on your
calculator.
๐๐๐๐๐ฅ = ๐๐๐ฅ
We can compare the graphs of ๐ฆ = ๐๐ฅ and ๐ฆ = ๐๐๐ฅ
14G
y = ex
y = lnx
y = x
(0,1)
(1,0)
Exponentials and Logarithms
The graph of ๐ = ๐๐๐ is a reflection of the graph ๐ = ๐๐ in
the line ๐ = ๐.
You should use ๐๐ when solving exponential equations involving ๐.
Solve the equation ๐๐ฅ = 5
14G
๐๐ฅ = 5
๐๐๐๐๐๐ฅ = ๐๐๐๐5
๐๐๐๐๐ฅ = ๐๐๐ฅ
๐ฅ = ๐๐๐๐5
๐ฅ๐๐๐๐๐ = ๐๐๐๐5
Take natural logs of both sides
Use the power law
๐๐๐๐๐ = 1
It is fine to leave your answers as exact values
๐๐ฅ = 5
๐๐๐๐ฅ = ๐๐5
๐ฅ = ๐๐5
Take natural logs of both sides
Use the power law
You should use notation like the above when dealing with natural logarithms
Exponentials and Logarithms
The graph of ๐ = ๐๐๐ is a reflection of the graph ๐ = ๐๐ in
the line ๐ = ๐.
You should use ๐๐ when solving exponential equations involving ๐.
Solve the equation ๐๐๐ฅ = 3
14G
๐๐๐ฅ = 3
๐๐๐๐๐ฅ = ๐๐๐ฅ
Think about what ln represents
๐๐๐๐๐ฅ = 3
๐3 = ๐ฅ
Rewrite without using a logarithm
๐๐๐ฅ = 3Learn this pattern so you can use it !
๐3 = ๐ฅ
Exponentials and Logarithms
The graph of ๐ = ๐๐๐ is a reflection of the graph ๐ = ๐๐ in
the line ๐ = ๐.
You should use ๐๐ when solving exponential equations involving ๐.
Solve the equation ๐2๐ฅ+3 = 7
14G
๐๐๐๐๐ฅ = ๐๐๐ฅ
๐2๐ฅ+3 = 7
2๐ฅ + 3 = ๐๐7
2๐ฅ = ๐๐7 โ 3
๐ฅ =1
2๐๐7 โ
3
2
Take natural logs
Divide by 2
Subtract 3
Exponentials and Logarithms
The graph of ๐ = ๐๐๐ is a reflection of the graph ๐ = ๐๐ in
the line ๐ = ๐.
You should use ๐๐ when solving exponential equations involving ๐.
Solve the equation 2๐๐๐ฅ + 1 = 5
14G
๐๐๐๐๐ฅ = ๐๐๐ฅ
2๐๐๐ฅ + 1 = 5Subtract 1
Write without a logarithm
Divide by 22๐๐๐ฅ = 4
๐๐๐ฅ = 2
๐2 = ๐ฅ
Exponentials and Logarithms
The graph of ๐ = ๐๐๐ is a reflection of the graph ๐ = ๐๐ in
the line ๐ = ๐.
You should use ๐๐ when solving exponential equations involving ๐.
Solve the equation ๐2๐ฅ + 5๐๐ฅ = 14
14G
๐๐๐๐๐ฅ = ๐๐๐ฅ
๐2๐ฅ + 5๐๐ฅ = 14
๐2๐ฅ + 5๐๐ฅ โ 14 = 0
๐๐ฅ + 7 ๐๐ฅ โ 2 = 0
๐๐ฅ = โ7 or ๐๐ฅ = 2
Subtract 14
๐๐ฅ = 2
No solutions if negative
๐ฅ = ๐๐2
Factorise (or write using y to help)
2 possible solutions for ๐๐ฅ
Take natural logs
Exponentials and Logarithms
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
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
In short, using a logarithmic scale can turn a curved line into a straight line!
14H
๐
๐ = ๐๐๐ ๐๐๐๐ = ๐๐๐๐๐๐๐๐๐
๐๐๐๐
These are the same equation, one with logs having been taken
If it is plotted on logarithmic axes, it is a straight line
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
This can be shown by starting with an exponential relationshipโฆ
14H
๐ฆ = ๐๐ฅ๐
๐๐๐๐ฆ = ๐๐๐๐๐ฅ๐
๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐ฅ๐
๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
๐ฆ = ๐๐ฅ + ๐
Take logs of both sides
Separate using the addition law
Use the power law
We can compare this with the straight line
formโฆThe โ๐ฆโ term is variable
We then have a constant multiplied by a variable
(๐ would be the constant)
We then have a constant
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
The data shows the rank (by size) and population of some UK cities.
The relationship between ๐ and ๐ can be modelled by the formula:
๐ = ๐๐ ๐
Where ๐ and ๐ are constants.
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
City Birmingham Leeds Glasgow Sheffield Bradford
Rank, ๐น 2 3 4 5 6
Population, ๐ท
1,000,000 730,000 620,000 530,000 480,000
A scatter graph of this data would look like this!
๐น
๐ท
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
The data shows the rank (by size) and population of some UK cities.
The relationship between ๐ and ๐ can be modelled by the formula:
๐ = ๐๐ ๐
Where ๐ and ๐ are constants.
a) Draw a table giving values of ๐๐๐๐ and ๐๐๐๐ to 2 decimal places
b) Plot a graph of ๐๐๐๐ against ๐๐๐๐ using the values from your table, and draw a
line of best fit
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
City Birmingham Leeds Glasgow Sheffield Bradford
Rank, ๐น 2 3 4 5 6
Population, ๐ท
1,000,000 730,000 620,000 530,000 480,000
City Birmingham Leeds Glasgow Sheffield Bradford
๐๐๐๐น
๐๐๐๐ท
0.30 0.48 0.60 0.70 0.78
6 5.86 5.79 5.72 5.68
๐๐๐๐น
๐๐๐๐ท
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
๐น
๐๐๐๐ท
๐๐๐๐น
๐ท
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
The data shows the rank (by size) and population of some UK cities.
The relationship between ๐ and ๐ can be modelled by the formula:
๐ = ๐๐ ๐
Where ๐ and ๐ are constants.
b) Plot a graph of ๐๐๐๐ against ๐๐๐๐ using the values from your table, and draw a
line of best fit
c) Use your graph to estimate the values of ๐ and ๐ to two significant figures
First you need to rearrange the original relationship into the
๐ฆ = ๐๐ฅ + ๐ form.
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
City Birmingham Leeds Glasgow Sheffield Bradford
๐๐๐๐น
๐๐๐๐ท
0.30 0.48 0.60 0.70 0.78
6 5.86 5.79 5.72 5.68
๐๐๐๐ท
๐๐๐๐น
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
The data shows the rank (by size) and population of some UK cities.
The relationship between ๐ and ๐ can be modelled by the formula:
๐ = ๐๐ ๐
Where ๐ and ๐ are constants.
c) Use your graph to estimate the values of ๐ and ๐ to two significant figures
First you need to rearrange the original relationship into the
๐ฆ = ๐๐ฅ + ๐ form.
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
๐ = ๐๐ ๐
๐๐๐๐ = ๐๐๐๐๐ ๐
๐๐๐๐ = ๐๐๐๐ + ๐๐๐๐ ๐
๐๐๐๐ = ๐๐๐๐ + ๐๐๐๐๐
Take logs
Addition law
Power law
So ๐๐๐๐ is the constant from the linear graph
And ๐ is the gradient from the linear graph
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
The data shows the rank (by size) and population of some UK cities.
The relationship between ๐ and ๐ can be modelled by the formula:
๐ = ๐๐ ๐
Where ๐ and ๐ are constants.
c) Use your graph to estimate the values of ๐ and ๐ to two significant figures
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
So ๐๐๐๐ is the constant from the linear graph
And ๐ is the gradient from the linear graph
๐๐๐๐ท
๐๐๐๐น
๐๐๐๐ = ๐. ๐
๐ = โ๐. ๐๐
๐ = ๐, ๐๐๐, ๐๐๐
๐ = ๐๐ ๐ ๐ = 1600000๐ โ0.67
๐๐๐.๐ = ๐
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
๐ = 1600000๐ โ0.67
๐น
๐ท
๐น
๐ท
Original data Calculated graph
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
There is a second situation that you may need to deal with, which is slightly
differentโฆ
๐ฆ = ๐๐๐ฅ
In this case, the variable x is in the power, rather than being raised to a
powerโฆ
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
๐ฆ = ๐๐๐ฅ
๐๐๐๐ฆ = ๐๐๐๐๐๐ฅ
๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
๐๐๐๐ฆ = ๐๐๐๐ + ๐ฅ๐๐๐๐
๐ฆ = ๐๐ฅ + ๐
Take logs of both sides
Separate using the addition law
Use the power law
We can compare this with the straight line
formโฆThe โ๐ฆโ term is variable
We then have a constant multiplied
by a variable(๐ would be the
constant)
We then have a constant
๐ฆ = ๐๐๐ฅ ๐๐๐๐ฆ = ๐๐๐๐ + ๐ฅ๐๐๐๐
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
Note that the axes we use for these will be different to the previous exampleโฆ
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ ๐ฆ = ๐๐๐ฅ ๐๐๐๐ฆ = ๐๐๐๐ + ๐ฅ๐๐๐๐
๐ฆ = ๐๐ฅ๐
๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ
๐ฆ = ๐๐๐ฅ
๐๐๐๐ฆ = ๐๐๐๐ + ๐ฅ๐๐๐๐
In this example, to get a linear relationship we plot
๐๐๐๐ฆ against ๐๐๐๐ฅ
In this example, to get a linear relationship we plot
๐๐๐๐ฆ against ๐ฅ
Ensure you check what form the equation is, as this will affect what you plot on the x-axisโฆ
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
The graph shown represents the growth of a population of bacteria, ๐ over a period of ๐ก 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 ๐ = ๐๐๐ก,
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 ๐ and ๐, giving them to 3sf where necessary
c) Interpret the meaning of the constant ๐ in this model
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ ๐ฆ = ๐๐๐ฅ ๐๐๐๐ฆ = ๐๐๐๐ + ๐ฅ๐๐๐๐
๐๐๐๐
๐ก
๐
๐
Use the form above as a starting pointโฆ
๐๐๐๐ฆ = ๐๐๐๐ + ๐ฅ๐๐๐๐
๐๐๐๐ = ๐๐๐๐ + ๐ก๐๐๐๐
๐๐๐๐ = 2 + 0.6๐ก
Use the variables in the question
The y-intercept is ๐๐๐๐ and the gradient is ๐๐๐๐
๐๐๐๐ = 2 + 0.6๐ก
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
The graph shown represents the growth of a population of bacteria, ๐ over a period of ๐ก 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 ๐ = ๐๐๐ก,
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 ๐ and ๐, giving them to 3sf where necessary
c) Interpret the meaning of the constant ๐ in this model
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ ๐ฆ = ๐๐๐ฅ ๐๐๐๐ฆ = ๐๐๐๐ + ๐ฅ๐๐๐๐
๐๐๐๐
๐ก
๐
๐
๐๐๐๐ = 2 + 0.6๐ก
๐ = 102+0.6๐ก
๐๐๐๐ = 2 + 0.6๐ก ๐ = 102 ร 100.6๐ก
๐ = 102 ร 100.6 ๐ก
๐ = 100 ร 3.98๐ก
๐ = 100 ๐ = 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 earlierโฆ
๐ = 100 ร 3.98๐ก
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
The graph shown represents the growth of a population of bacteria, ๐ over a period of ๐ก 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 ๐ = ๐๐๐ก,
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 ๐ and ๐, giving them to 3sf where necessary
c) Interpret the meaning of the constant ๐ in this model
14H
๐ฆ = ๐๐ฅ๐ ๐๐๐๐ฆ = ๐๐๐๐ + ๐๐๐๐๐ฅ ๐ฆ = ๐๐๐ฅ ๐๐๐๐ฆ = ๐๐๐๐ + ๐ฅ๐๐๐๐
๐๐๐๐
๐ก
๐
๐
๐๐๐๐ = 2 + 0.6๐ก
๐ = 100 ร 3.98๐ก
๐ = 100 ร 3.98๐ก
๐ = 100
Let ๐ก = 0
Calculate
๐ = 100 ร 3.980
So in this case, 100 gives the initial size of the bacteria population