C2: Chapter 3 Logarithms
Last modified: 1 st September 2015

Sketch a graph of y = 3 x
(Note: this was once used as an exam question)
1
?
1 mark: Two of the three criteria.
2 marks: All of the three criteria.
(a) Correct shape in left quadrant.
(b) Correct shape in right quadrant
(c) y-intercept of 1.

Some operators exist to provide the opposite of others.
4
4
Function
4
4
2
5
12
7
16
1024
Inverse

5
4
4
4
4

Some operators exist to provide the opposite of others.
Function Inverse
81
4
x
?
3
4
How therefore would describe the effect of log
3 x in words?
log
3 x
We describe this as the “logarithm of x base 3” or “log of x base 3” or
“taking the log of x base 3”.
It finds the power that, when 3 is raised to it, gives you x.
i.e. If y = log
3 x, then 3 y = x
It is the opposite/inverse of exponentiation.

Remember that logarithms find the missing power.
Bro Tip #1: Imagine what power would slot in the middle of the two.
log
2
8 = 3 ?
log
3
9 = 2 ?
log
10
100 = 2
?
log
4
1 = 0
?
log
3
3 = 1 ?
log
2
Bro Tip #2: log a
1 = 0 (a > 0)
Bro Tip #3: log a a = 1 (a>0)
Click to start bromanimation

Remember that logarithms find the missing power.
log
2
( ) = -1
2
?
log
2
( ) = -3
8
?
log
3
( ) = -4
81
?
log
4
(-1) =
i__ log
?
e
4
Bro Tip #4: When we take the log of any value between 0 and 1 (exclusive), we end up with a negative number.
Bro Tip #5: If you want a real result, you can only take logs of positive values.

x y
0.25
0.5
8
1 2 4 8
-2
?
-1
?
0
? ? ?
3
?
6
4
2 y = log
2 x
-4 -2 2 4 6 8 10 12
-2
14
-4
-6
Click to brosketch

Logs help us solve equations when the power is unknown.
Find the x for which 10 x = 500
We can write this as x = log
10
Broculator Tip:
The log button on your calculator is implicitly base 10. So “[log] [500]” will give you log
10
500

2 3 = 8
9 2 = 81
3 4 = 81
3 x = 55
Bro Tip: In both cases, the 2 is the ‘base’.
log
2
8 = 3 log
9
81 = 2 log
3
81 = 4 log
3
55 = x

C2 Chapter 3 Pg 42
Exercise 3B (All questions)
Exercise 3C (Q1, 3, 5, 7)

These are 3 laws of logs that you need to remember. log a xy = log a x + log a y log a x y
= log a x - log a y log a
(x k ) = k log a x
Proving these involves rewriting the logs as exponential expressions, then using laws of indices.

Write the following as a single logarithm.
log
2
30 – log
2
6
2log
3 a + log
3 b
3log
4
(a) – 4log
4 b
3log a
(a
b) log
2
?
5 log
3
(a 2 b) log
4
?
b 4
3
( )
Put in the form k + log a
(..)
?

Now the other way round! Write in the form log a x, log a y and log a z.
log a
(b 2 c 3 ) 2log a b + 3log a c
4log a
(
b) log log a a
(a
b)
2log a b ?
1 + log a b
-log a x
?

The well-known “Moore’s Law” states that the processing power of computers doubles every two years. It’s been remarkably accurate so far!
If we were to plot the number of transistors against
The graph would look rubbish if we chose a range of values on the y-axis to accommodate all the values, because except for the last few years, most of the points would look close to 0.

The well-known “Moore’s Law” states that the processing power of computers doubles every two years. It’s been remarkably accurate so far!
This graph gets around the problem by letting the y-values increase by a factor of 10 for each unit, rather than increasing by a constant amount each time. Technically this is not allowed!

The well-known “Moore’s Law” states that the processing power of computers doubles every two years. It’s been remarkably accurate so far!
10
9
We could instead use a logarithmic scale. We can take the log base 10 of these values. Then we’ll get 3, 4, 5, 6, ..., which is now allowed!*
8
7
6
5
4
Logarithmic scales turn exponential graphs into linear ones (i.e. a straight line), thus making it much easier to plot all the points together.
3
* Although realistically, a scale of 10, 100, 1000, etc. is permissible as long as we’re mindful that it’s a logarithmic scale.

Logarithmic scales are used for earthquakes and noise levels.
log a ax = 1 + log a x
From our laws of logs, in base a... when a quantity gets a times bigger, the overall result only increases by 1.
Thus using logarithms turns a factor difference into a constant difference.
The Richter Scale is used to measure the magnitude of earthquakes. The scale is logarithmic (base 10): it means if amplitude of the earthquake’s waves gets 10 times bigger, the value on the Richter Scale only increases by 1.
Earthquakes of magnitude 6 vs 7 doesn’t look like a substantial difference, but just the one point difference means it’s ten times worse!

C2 Chapter 3 Pg 42
Exercise 3D (All questions)

x
We saw how we can solve equations like 10 x = 125.
But what about when the base is different, e.g. 3 x = 20?
OPTION 1: The “Look at me, I have a fancy calculator” method x = log
3 ?
OPTION 2: The “change of base” method
3 x = 20 log
10 x log
10
3 x = log
10
3 = log
10 log
10 log
10
20
3
20
20
Super Bro Tip:
Whenever you’re trying to solve an equation where the variable appears in the power, your first instinct should always be TAKE
LOGS DAMMIT!

We saw a second ago that we could “change the base” to find log
3 in terms of log
10
.
3 x = 20
METHOD 1 x = log
3
20
METHOD 2 x = log
10 log
10
20
3
More generally, to change the base from a to b: log a a x log log b b x a
Click to start bromanimation

Express these logarithms in the specified new base.
log log log log
2
7
10
5
5
10
5
10 in base 10 in base 12 in base 9 in base 10 log log
10
5
2 log log
12
10
7 log log
9
?
9
5
10 log
10 log
10
10
5
Broculator Tip:
This is how you could find log
2
5 on a calculator if you didn’t have the fancy extra log button. i.e.
Change to base
10!
___1___ log
10
5

Bro Tip: When you switch the argument and base, you take the reciprocal.
log a b =
___
1___ log b a

7 x+1 = 3 x+2 log 7 x+1 = log 3 x+2
(x+1)log 7 = (x+2)log 3
Variables appear in powers, so apply Bro Tip.
(The base of the log doesn’t matter) xlog 7 + log 7 = xlog 3 + 2log 3 xlog 7 - xlog 3 = 2log 3 – log 7 x(log 7 - log 3) = 2log 3 – log 7 x =
2log 3 – log 7 log 7 – log 3

(5
5 2x + 7(5 x ) – 30 = 0 x ) 2 + 7(5 x ) – 30 = 0
Let y = 5 x y 2 + 7y – 30 = 0
(y+10)(y-3) = 0 y = -10 or y = 3
5 x = -10 or 5 x = 3 x = log
5
(-10) or x = log
5
3
Bro Tip:
By recognising that 5 2x = (5 x ) 2 , we’ve turned the equation into a quadratic!

2 2x + 3(2 x ) – 4 = 0 3 x-1 = 8 x+1 x = 0 ?

When solving, you can often either: a) Get in the form log a
b = c. Then rearrange as a c b) Get in the form log a b = log a
c. Then b = c.
= b
EdExcel exam questions:
Solve log
2
(2x + 1) – log
2
x = 2 Solve log
2
(x + 1) – log
2
x = log
2
7

When solving, you can often either: a) Get in the form log a
b = c. Then rearrange as a c b) Get in the form log a b = log a
c. Then b = c.
= b
Edexcel exam questions:
Solve log
2
32
 log
2
16
 log
2 x log
2 x log
2
(11 – 6x) = 2 log
2
(x – 1) + 3