1. Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com
C2: Chapter 3 Logarithms
Dr J Frost
Last modified: 1st September 2015

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

3. 4 4 4 4 4 4 4 4 Functions and their inverses x × 3 x  3 x + 3 x - 3
Some operators exist to provide the opposite of others.
Function
Inverse 12 4 4
x × 3
x  3 ? 7 4 4
x + 3 ?
x - 3 16 4 4 x2 ? √x
1024 4 4 x5 ?
5√x

4. log3 x 4 4 Functions and their inverses 3x log3 x 81 ?
Some operators exist to provide the opposite of others.
Function
Inverse 81 4 4 3x ?
log3 x
How therefore would describe the effect of log3 x in words?
log3 x
It finds the power that, when 3 is raised to it, gives you x.
i.e. If y = log3 x, then 3y = x
It is the opposite/inverse of exponentiation.
We describe this as the “logarithm of x base 3” or “log of x base 3” or “taking the log of x base 3”.

5. Click to start bromanimation
Computing logs
Remember that logarithms find the missing power.
Bro Tip #1: Imagine what power would slot in the middle of the two.
log = 3 ?
Click to start bromanimation
log = 3 3=
log = 2 ?
log = 2 ?
Bro Tip #2: loga 1 = 0 (a > 0)
log = 0 ?
Bro Tip #3: loga a = 1 (a>0)
log = 1 ?

6. log2( ) = -1 log2( ) = -3 log3( ) = -4 log4 (-1) = 1 2 1 8 1 81 __πi__
Computing logs
Remember that logarithms find the missing power.
log2( ) = -1 1 2
Bro Tip #4: When we take the log of any value between 0 and 1 (exclusive), we end up with a negative number. ? 1 8
log2( ) = -3 ?
log3( ) = -4 1 81 ?
Bro Tip #5: If you want a real result, you can only take logs of positive values.
__πi__
loge 4 ?
log4 (-1) =

7. 8 6 4 2 -2 -4 -6
y = log2 x x
0.25
0.5 1 2 4 8 y -2 -1 3 ? ? ? ? ? ?
Get students to sketch axes and tables in their books.
Click to brosketch

8. Using logs Find the x for which 10x = 500
Logs help us solve equations when the power is unknown.
Find the x for which 10x = 500
We can write this as x = log10 500 ?
Broculator Tip:
The log button on your calculator is implicitly base 10. So “[log] [500]” will give you log10 500

9. 23 = 8 log2 8 = 3 log9 81 = 2 92 = 81 log3 81 = 4 34 = 81 log3 55 = x
More on rewriting Powers as Logarithms
Bro Tip: In both cases, the 2 is the ‘base’.
23 = 8
log = 3 ?
log = 2
92 = 81
log = 4 ?
34 = 81 ?
log = x
3x = 55

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

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

12. log230 – log26 log2 5 2log3 a + log3 b log3 (a2b) log4 ( )
Laws of Logs
Write the following as a single logarithm.
log230 – log26
log2 5 ?
2log3 a + log3 b
log3 (a2b) ? a3 b4
3log4 (a) – 4log4 b
log4 ( ) ?
Put in the form k + loga (..) ?
3loga (a√b)

13. loga(b2c3) 2loga b + 3logac 4loga(√b) 2loga b loga(a√b) 1 + loga b
Laws of Logs
Now the other way round! Write in the form loga x, loga y and loga z.
loga(b2c3)
2loga b + 3logac ?
4loga(√b)
2loga b ? 1 2
loga(a√b) ?
1 + loga b 1 x
loga( )
-loga x ?

14. Using logs in science
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 year, the type of graph would be exponential.
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. ?

15. Using logs in science
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!

16. log10(Number of transistors)
Using logs in science
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
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!* 9 8 7
log10(Number of transistors) 6 5
Logarithmic scales turn exponential graphs into linear ones (i.e. a straight line), thus making it much easier to plot all the points together. 4 3
* Although realistically, a scale of 10, 100, 1000, etc. is permissible as long as we’re mindful that it’s a logarithmic scale.

17. loga ax = 1 + loga x Using logs in science
Logarithmic scales are used for earthquakes and noise levels.
loga ax = 1 + loga 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!

18. Exercise 3D (All questions)
Exercises
C2 Chapter 3 Pg 42
Exercise 3D (All questions)

19. x = log3 20 3x = 20 log10 3x = log10 20 x log10 3 = log10 20 x =
Solving ax = b
We saw how we can solve equations like 10x = 125.
But what about when the base is different, e.g. 3x = 20?
OPTION 1: The “Look at me, I have a fancy calculator” method
x = log3 20 ? ?
OPTION 2: The “change of base” method
3x = 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!
log10 3x = log10 20 ?
x log10 3 = log10 20 ? ?
x =
log10 20
log10 3

20. Click to start bromanimation
Changing the Base
We saw a second ago that we could “change the base” to find log3 in terms of log10.
3x = 20
METHOD 1
METHOD 2
x = log3 20
x =
log10 20
log10 3
More generally, to change the base from a to b:
logb x
logb a
loga x = x
Click to start bromanimation a

21. log2 5 in base 10 log7 10 in base 12 log10 5 in base 9
All your base belong to us
Express these logarithms in the specified new base.
log2 5 in base 10
log10 5
log10 2 ?
Broculator Tip:
This is how you could find log25 on a calculator if you didn’t have the fancy extra log button. i.e. Change to base 10!
log12 10
log12 7 ?
log7 10 in base 12
log10 5 in base 9
log9 5
log9 10 ?
log5 10 in base 10
log10 10
log10 5 ?
___1___
log10 5 =

22. loga b = logb a All your base belong to us ___1___
Bro Tip: When you switch the argument and base, you take the reciprocal.
___1___
logb a
loga b =

23. 7x+1 = 3x+2 log 7x+1 = log 3x+2 (x+1)log 7 = (x+2)log 3
Solving Equations involving Variables in Powers
7x+1 = 3x+2
Variables appear in powers, so apply Bro Tip.
(The base of the log doesn’t matter)
log 7x+1 = log 3x+2
(x+1)log 7 = (x+2)log 3
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

24. 52x + 7(5x) – 30 = 0 (5x)2 + 7(5x) – 30 = 0 Let y = 5x
Solving Equations involving Variables in Powers
52x + 7(5x) – 30 = 0
Bro Tip:
By recognising that 52x = (5x)2, we’ve turned the equation into a quadratic!
(5x)2 + 7(5x) – 30 = 0
Let y = 5x
y2 + 7y – 30 = 0
(y+10)(y-3) = 0
y = -10 or y = 3
Can’t have log of a negative number, so not a real solution.
5x = -10 or 5x = 3
x = log5(-10) or x = log53

25. 22x + 3(2x) – 4 = 0
3x-1 = 8x+1
x = 0 ?
x = -3.24 ?

26. Exam Technique ? ? Solve log2 (2x + 1) – log2 x = 2
When solving, you can often either:
Get in the form logab = c. Then rearrange as ac = b
Get in the form logab = logac. Then b = c.
EdExcel exam questions:
Solve log2 (2x + 1) – log2 x = 2
Solve log 2 (x + 1) – log 2 x = log 2 7
Answer: x = 1/2 ?
Answer: x = 1/6 ?

27. Exam Technique ? ? log2 (11 – 6x) = 2 log2 (x – 1) + 3
When solving, you can often either:
Get in the form logab = c. Then rearrange as ac = b
Get in the form logab = logac. Then b = c.
Edexcel exam questions:
log2 (11 – 6x) = 2 log2 (x – 1) + 3
Solve ? ?
Answer: x = 8 or 1/8
Answer: x = -1/4 or 3/2