Aaron Walcott, TT
What Is a Logarithm?
A logarithm answers the question:
“To what power must the base be raised to get this number?”
If: 23 = 8
Then: log2 8 = 3
Logs are simply exponents done backwards.
They allow us to:
• Break apart multiplication
• Simplify division
• Bring powers down in front
They are especially useful in algebra, calculus, growth/decay, and exam questions involving
unknown exponents.
Worked Example 1 (Product Law)
Simplify: log2 (8𝑥)
Split using product rule: = log2 8 + log2 𝑥
= 3 + log2 𝑥
Worked Example 2 (Power Law)
Simplify: log3 (92 )
Apply power law: = 2log3 9
=2×2=4
1
, Aaron Walcott, TT
Logarithm Pinch Points
• Forgetting that logs turn multiplication into addition
• Losing minus signs in quotient rule
• Forgetting log of 1 = 0
• Forgetting log of base = 1
• Not recognising when change-of-base is required
• Treating log (a + b) as log a + log b (this is incorrect)
LAWS OF LOGARITHMS
Product Law log𝑎 (𝑥𝑦) = log𝑎 𝑥 + log𝑎 𝑦
𝑥
Quotient Law log𝑎 (𝑦) = log𝑎 𝑥 − log𝑎 𝑦
Power Law log𝑎 (𝑥 𝑘 ) = 𝑘log𝑎 𝑥
Log of 1 log𝑎 1 = 0
Log of Base log𝑎 𝑎 = 1
2
What Is a Logarithm?
A logarithm answers the question:
“To what power must the base be raised to get this number?”
If: 23 = 8
Then: log2 8 = 3
Logs are simply exponents done backwards.
They allow us to:
• Break apart multiplication
• Simplify division
• Bring powers down in front
They are especially useful in algebra, calculus, growth/decay, and exam questions involving
unknown exponents.
Worked Example 1 (Product Law)
Simplify: log2 (8𝑥)
Split using product rule: = log2 8 + log2 𝑥
= 3 + log2 𝑥
Worked Example 2 (Power Law)
Simplify: log3 (92 )
Apply power law: = 2log3 9
=2×2=4
1
, Aaron Walcott, TT
Logarithm Pinch Points
• Forgetting that logs turn multiplication into addition
• Losing minus signs in quotient rule
• Forgetting log of 1 = 0
• Forgetting log of base = 1
• Not recognising when change-of-base is required
• Treating log (a + b) as log a + log b (this is incorrect)
LAWS OF LOGARITHMS
Product Law log𝑎 (𝑥𝑦) = log𝑎 𝑥 + log𝑎 𝑦
𝑥
Quotient Law log𝑎 (𝑦) = log𝑎 𝑥 − log𝑎 𝑦
Power Law log𝑎 (𝑥 𝑘 ) = 𝑘log𝑎 𝑥
Log of 1 log𝑎 1 = 0
Log of Base log𝑎 𝑎 = 1
2