Definition of logarithms, laws of logs, change of base formula.
Answer at least 3 of 3 correctly to complete this section.
The Richter scale, decibels, and pH all use logarithms. Why? Because some quantities vary so enormously — from tiny earthquakes to devastating ones — that a linear scale is useless. Logarithms compress huge ranges into manageable numbers.
At A-Level, logarithms are the tool that lets you solve exponential equations. If you know that 2x=32, you can spot that x=5. But what if 3x=20? You cannot solve that by inspection — you need logarithms.
A logarithm answers a single question: “what power do I need?”
loga(x)=nmeansan=x
Read log2(8)=3 as: “the power of 2 that gives 8 is 3”, because 23=8.
The logarithm is the inverse of exponentiation — it undoes what a power does. If exponentiation asks “what do I get when I raise a to the power n?”, then the logarithm asks “what power of a gives me x?”
Every exponential equation has an equivalent logarithmic form, and vice versa:
| Exponential form | Logarithmic form |
|---|---|
| 23=8 | log2(8)=3 |
| 102=100 | log10(100)=2 |
| 5−1=51 | log5(51)=−1 |
| 30=1 | log3(1)=0 |
| 41/2=2 | log4(2)=21 |
Notice the pattern: the base stays the same, but the exponent and the result swap roles.
See how exponential and logarithmic forms relate
Important restrictions: You can only take the logarithm of a positive number. log2(0) and log2(−4) are undefined. The base must be positive and not equal to 1.
Answer at least 3 of 4 correctly to complete this section.
The three log laws follow directly from the index laws. If you know your index laws, you already know your log laws — the operations just shift.
| Index law | Log law | What happens |
|---|---|---|
| am×an=am+n | loga(xy)=logax+logay | Multiplication becomes addition |
| am÷an=am−n | loga(yx)=logax−logay | Division becomes subtraction |
| (am)n=amn | loga(xn)=nlogax | Powers come out as multipliers |
loga(xy)=logax+logay
“The log of a product equals the sum of the logs.”
Example: log2(4×8)=log2(4)+log2(8)=2+3=5
Check: 4×8=32 and log2(32)=5. It works.
loga(yx)=logax−logay
“The log of a quotient equals the difference of the logs.”
Example: log3(27)−log3(9)=3−2=1
Check: 927=3 and log3(3)=1. It works.
loga(xn)=nlogax
“The log of a power brings the exponent out front as a multiplier.”
Example: log2(82)=2log2(8)=2×3=6
Check: 82=64 and log2(64)=6. It works.
Drag the sliders to verify each law with real numbers
The classic trap: log(a+b)=log(a)+log(b). The product law says log(a⋅b)=log(a)+log(b). There is no law for the log of a sum. Students who confuse these lose marks every year.
Answer at least 3 of 4 correctly to complete this section.
These follow directly from the definition and come up constantly:
loga(a)=1because a1=a
loga(1)=0because a0=1
loga(an)=nbecause the log undoes the power
aloga(x)=xbecause the power undoes the log
The last two show that loga and a(⋅) are inverse functions — each undoes the other.
Your calculator has buttons for log10 and ln (which is loge), but not for log2 or log5. The change of base formula lets you convert:
logax=logbalogbx
You can choose any base b for the conversion — 10 and e are the most common since your calculator has them.
Example: Find log3(20) using your calculator.
log3(20)=log10(3)log10(20)=0.4771.301=2.727 (3 d.p.)
Check: 32.727≈20. It works.
Convention: At A-Level, log without a subscript means log10, and ln means loge (the natural logarithm). Always check whether the question specifies a base.
Answer at least 3 of 4 correctly to complete this section.
There are two main strategies for solving equations with logarithms:
If you have loga(something)=n, rewrite as an=something.
Example: Solve log2(x)=5.
log2(x)=5⟹25=x⟹x=32
If you have multiple log terms, combine them into a single log using the laws, then convert.
Example: Solve log3(x)+log3(4)=log3(20).
Using the product law on the left:
log3(4x)=log3(20)
Since the logs are equal (same base), the arguments must be equal:
4x=20⟹x=5
Solve 2log5(x)−log5(3)=log5(12).
Step 1: Use the power law on 2log5(x):
log5(x2)−log5(3)=log5(12)
Step 2: Use the quotient law:
log5(3x2)=log5(12)
Step 3: Arguments must be equal:
3x2=12⟹x2=36⟹x=6
We take x=6 (not x=−6) because we need log5(x) to be defined, and you cannot take the log of a negative number.
Solve 3x=20.
Take log10 of both sides:
log(3x)=log(20)
Use the power law:
xlog(3)=log(20)
x=log(3)log(20)=0.4771.301=2.73 (3 s.f.)
Always check: After solving, substitute back to make sure you are not taking the log of zero or a negative number. If x=−6 appeared as a solution, you would reject it because log5(−6) is undefined.
Answer at least 3 of 4 correctly to complete this section.
Write 2loga(3)+loga(4) as a single logarithm.
2loga(3)=loga(32)=loga(9)
loga(9)+loga(4)=loga(9×4)=loga(36)
Solve 5x=12, giving your answer to 3 significant figures.
log(5x)=log(12) xlog(5)=log(12) x=log(5)log(12)=0.6991.079=1.54 (3 s.f.)
Solve log2(x+1)+log2(x−1)=3.
Product law:
log2((x+1)(x−1))=3
log2(x2−1)=3
Convert to exponential form:
x2−1=23=8
x2=9⟹x=3
We reject x=−3 because log2(x−1) requires x>1, and −3<1.
| Pattern | Technique |
|---|---|
| loga(x)=n | Convert: x=an |
| loga(f)+loga(g)=loga(k) | Product law, then equate arguments |
| loga(f)−loga(g)=loga(k) | Quotient law, then equate arguments |
| nloga(x)=loga(k) | Power law, then equate arguments |
| ax=k | Take logs of both sides, use power law |
Answer at least 3 of 4 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
Simplifying surds, laws of indices, rationalising denominators.
Using logarithms to solve exponential equations. Graphs of exponential functions.
The natural logarithm and exponential function. Properties of e^x and ln x.