Twitter: @Owen134866
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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
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 smallest…
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 £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…)
Exponentials and Logarithms
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!
Exponentials and Logarithms
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 = x3
So dy/dx = 3x2
y = x3y = 3x2
The Gradient at this point…
… 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 function…
We can plot the gradient function on the graph itself…
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 function…
We can plot the gradient function on the graph itself…
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 crossed…
Therefore the must be a value between 2 and 3 where the lines are equal…
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 conclusion…
If y = ex
Then dy/dx = ex as well!
y = ex
y = ex
Exponentials and Logarithms
14B
Exponentials and Logarithms
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!
Exponentials and Logarithms
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)
Exponentials and Logarithms
14B
If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥
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
𝒚 = 𝒆𝒙
y = ex
y = ex + 2
14B
If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥
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) ‘swap signs’
You need to be able to sketch transformations of the graph
𝒚 = 𝒆𝒙
y = ex
y = -ex
14B
If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥
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
𝒚 = 𝒆𝒙
y = ex
y = e2x
14B
If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥
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 before…)
(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
𝒚 = 𝒆𝒙
y = ex
y = ex + 1
14B
If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥
Exponentials and Logarithms
y = ex
The graph of e-x, but with y values 10
times bigger…
y = e-x
y = 10e-x
(0, 1)
(0, 10)
You need to be able to sketch transformations of the graph
𝒚 = 𝒆𝒙
y = ex
y = 10e-x
14B
If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥
Exponentials and Logarithms
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)
You need to be able to sketch transformations of the graph
𝒚 = 𝒆𝒙
y = ex
y = 3 + 4e0.5x
14B
If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥
Exponentials and Logarithms
Exponentials and Logarithms
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
If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥
Exponentials and Logarithms
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
If: 𝑓 𝑥 = 𝑒𝑥Then: 𝑓′ 𝑥 = 𝑒𝑥 If: 𝑓 𝑥 = 𝑒𝑘𝑥Then: 𝑓′ 𝑥 = 𝑘𝑒𝑘𝑥
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
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 ‘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’
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:
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)
Exponentials and LogarithmsLaws of logarithms
Solve the equation:
𝑙𝑜𝑔104 + 2𝑙𝑜𝑔10𝑥 = 2
log log loga a axy x y log log loga a a
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…
Exponentials and LogarithmsLaws of logarithms
Solve the equation:
𝑙𝑜𝑔3 𝑥 + 11 − 𝑙𝑜𝑔3 𝑥 − 5 = 2
14E
log log loga a axy x y log log loga a a
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
Exponentials and LogarithmsSolving Equations using Logarithms
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
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 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
Exponentials and Logarithms
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)
Exponentials and Logarithms
The graph of 𝒚 = 𝒍𝒏𝒙 is a reflection of the graph 𝒚 = 𝒆𝒙 in
the line 𝒚 = 𝒙.
You should use 𝑙𝑛 when solving exponential equations involving 𝑒.
Solve the equation 𝑒𝑥 = 5
14G
𝑒𝑥 = 5
𝑙𝑜𝑔𝑒𝑒𝑥 = 𝑙𝑜𝑔𝑒5
𝑙𝑜𝑔𝑒𝑥 = 𝑙𝑛𝑥
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
Exponentials and Logarithms
The graph of 𝒚 = 𝒍𝒏𝒙 is a reflection of the graph 𝒚 = 𝒆𝒙 in
the line 𝒚 = 𝒙.
You should use 𝑙𝑛 when solving exponential equations involving 𝑒.
Solve the equation 𝑙𝑛𝑥 = 3
14G
𝑙𝑛𝑥 = 3
𝑙𝑜𝑔𝑒𝑥 = 𝑙𝑛𝑥
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
Exponentials and Logarithms
The graph of 𝒚 = 𝒍𝒏𝒙 is a reflection of the graph 𝒚 = 𝒆𝒙 in
the line 𝒚 = 𝒙.
You should use 𝑙𝑛 when solving exponential equations involving 𝑒.
Solve the equation 𝑒2𝑥+3 = 7
14G
𝑙𝑜𝑔𝑒𝑥 = 𝑙𝑛𝑥
𝑒2𝑥+3 = 7
2𝑥 + 3 = 𝑙𝑛7
2𝑥 = 𝑙𝑛7 − 3
𝑥 =1
2𝑙𝑛7 −
3
2
Take natural logs
Divide by 2
Subtract 3
Exponentials and Logarithms
The graph of 𝒚 = 𝒍𝒏𝒙 is a reflection of the graph 𝒚 = 𝒆𝒙 in
the line 𝒚 = 𝒙.
You should use 𝑙𝑛 when solving exponential equations involving 𝑒.
Solve the equation 2𝑙𝑛𝑥 + 1 = 5
14G
𝑙𝑜𝑔𝑒𝑥 = 𝑙𝑛𝑥
2𝑙𝑛𝑥 + 1 = 5Subtract 1
Write without a logarithm
Divide by 22𝑙𝑛𝑥 = 4
𝑙𝑛𝑥 = 2
𝑒2 = 𝑥
Exponentials and Logarithms
The graph of 𝒚 = 𝒍𝒏𝒙 is a reflection of the graph 𝒚 = 𝒆𝒙 in
the line 𝒚 = 𝒙.
You should use 𝑙𝑛 when solving exponential equations involving 𝑒.
Solve the equation 𝑒2𝑥 + 5𝑒𝑥 = 14
14G
𝑙𝑜𝑔𝑒𝑥 = 𝑙𝑛𝑥
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
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
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
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 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
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 𝑃 and 𝑅 can be modelled by the formula:
𝑃 = 𝑎𝑅𝑛
Where 𝑎 and 𝑛 are constants.
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥
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
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 𝑃 and 𝑅 can be modelled by the formula:
𝑃 = 𝑎𝑅𝑛
Where 𝑎 and 𝑛 are constants.
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
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥
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
City Birmingham Leeds Glasgow Sheffield Bradford
𝒍𝒐𝒈𝑹
𝒍𝒐𝒈𝑷
0.30 0.48 0.60 0.70 0.78
6 5.86 5.79 5.72 5.68
𝒍𝒐𝒈𝑹
𝒍𝒐𝒈𝑷
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥
𝑹
𝒍𝒐𝒈𝑷
𝒍𝒐𝒈𝑹
𝑷
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 𝑃 and 𝑅 can be modelled by the formula:
𝑃 = 𝑎𝑅𝑛
Where 𝑎 and 𝑛 are constants.
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.
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥
City Birmingham Leeds Glasgow Sheffield Bradford
𝒍𝒐𝒈𝑹
𝒍𝒐𝒈𝑷
0.30 0.48 0.60 0.70 0.78
6 5.86 5.79 5.72 5.68
𝒍𝒐𝒈𝑷
𝒍𝒐𝒈𝑹
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 𝑃 and 𝑅 can be modelled by the formula:
𝑃 = 𝑎𝑅𝑛
Where 𝑎 and 𝑛 are constants.
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.
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥
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
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 𝑃 and 𝑅 can be modelled by the formula:
𝑃 = 𝑎𝑅𝑛
Where 𝑎 and 𝑛 are constants.
c) Use your graph to estimate the values of 𝑎 and 𝑛 to two significant figures
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥
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
Exponentials and Logarithms
Logarithms can be used to manage and explore non-linear trends in data
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥
𝑃 = 1600000𝑅−0.67
𝑹
𝑷
𝑹
𝑷
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
different…
𝑦 = 𝑎𝑏𝑥
In this case, the variable x is in the power, rather than being raised to a
power…
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥
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
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 example…
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥 𝑦 = 𝑎𝑏𝑥 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑥𝑙𝑜𝑔𝑏
𝑦 = 𝑎𝑥𝑛
𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥
𝑦 = 𝑎𝑏𝑥
𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑥𝑙𝑜𝑔𝑏
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…
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, 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
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥 𝑦 = 𝑎𝑏𝑥 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑥𝑙𝑜𝑔𝑏
𝑙𝑜𝑔𝑃
𝑡
𝟐
𝑂
Use the form above as a starting point…
𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑥𝑙𝑜𝑔𝑏
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
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, 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
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥 𝑦 = 𝑎𝑏𝑥 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑥𝑙𝑜𝑔𝑏
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
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, 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
14H
𝑦 = 𝑎𝑥𝑛 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑛𝑙𝑜𝑔𝑥 𝑦 = 𝑎𝑏𝑥 𝑙𝑜𝑔𝑦 = 𝑙𝑜𝑔𝑎 + 𝑥𝑙𝑜𝑔𝑏
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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