01 · Definition
Rewrite 4³ = 64 using a logarithm.
Hint
The logarithm is the exponent 3.
Worked solution
log₄(64) = 3.
Understand · explore · practise
Logarithms and log laws
Understand what a logarithm means, calculate exact values, prove and use the product, quotient and power laws, change base and avoid domain errors.
Before you startIndex laws and positive-base exponentials
01 / A logarithm is an exponent
Ask which power produces the given number.
loga(N) = x ⇔ aˣ = N
a > 0, a ≠ 1, N > 0
For example, log₂(32) = 5 because 2⁵ = 32. The base is the number being raised to a power; the logarithm is the exponent you need.
5⁻³ = 1/125 ⇔ log₅(1/125) = −3
91/2 = 3 ⇔ log₉(3) = 1/2
The input must be positive, but the logarithm can be negative, zero or positive. Base 1 is excluded because 1ˣ is always 1 and cannot be inverted uniquely.
02 / Exact logarithms and special values
Rewrite the input using the base.
loga(1) = 0
loga(a) = 1
log₄(64) = 3
log₃(1/9) = −2
log₃₂(2) = 1/5
log1/3(27) = −3
A base between 0 and 1 is allowed. Its exponential decreases, so its logarithm decreases too. For base 2, 2⁵ < 50 < 2⁶ gives 5 < log₂(50) < 6. Always account for the direction when the base is below 1.
03 / The product law
Multiplication adds exponents.
For positive x and y, write x = aᵐ and y = aⁿ. Then xy = am+n, so:
loga(xy) = loga(x) + loga(y)
Choose a law and change n in the model. It holds the first exponent at 2 and lets you compare the power calculation with the logarithm calculation. No timer changes the working.
Compare a log law with its index lawMove at your pace
Product: multiply inputs. With first exponent 2 and n = 2, the resulting input is 16 and its base-2 logarithm is 4. Multiplying powers adds their exponents. All inputs are positive, including when n is negative.
Watch multiplication become addition of exponents
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / The quotient and power laws
Division subtracts exponents; a power multiplies them.
loga(x/y) = loga(x) − loga(y)
loga(xᵏ) = k loga(x)
Use the same valid base throughout and positive x,y. If x = aᵐ and y = aⁿ, then x/y = am−n. Also xᵏ = amk, which gives the power law for real k.
loga(1/x) = −loga(x)
loga(√x) = ½loga(x)
The subtraction is between the logarithms, not between their inputs. log(x) − log(y) becomes log(x/y), not log(x − y).
05 / Combine into one logarithm
Deal with coefficients before products and quotients.
Simplify 2log₃(2) + log₃(9) − log₃(4)Worked example
2log₃(2) = log₃(4)
Move the coefficient into an exponent.
log₃(4) + log₃(9) − log₃(4)
= log₃[(4 × 9)/4]
Addition multiplies inputs; subtraction divides them.
= log₃(9) = 2
Finish with an exact logarithm.
A fractional coefficient
log₂(40) − ½log₂(25) − ⅓log₂(8)
= log₂[40/(5 × 2)] = log₂(4) = 2
Roots appear when a fractional coefficient moves into the input as a power.
06 / Expand a logarithm
State positive-input assumptions before splitting.
For x,y,z > 0 and a valid base a:
loga(x³√y/z²)
= 3loga(x) + ½loga(y) − 2loga(z)
loga(a²x) = 2 + loga(x)
If p = logb(8), then logb(2) = p/3 and logb(4b) = 2p/3 + 1. Expressing all inputs in terms of familiar factors makes the given logarithm useful.
07 / Two tempting rules that are false
The laws do not split a sum inside a logarithm.
log₂(8 + 8) = log₂(16) = 4
log₂(8) + log₂(8) = 3 + 3 = 6
Therefore log(x + y) is not generally log(x) + log(y). The correct addition law starts with a product inside one logarithm.
Why log(x²) needs care
For x > 0, log(x²) = 2log(x). For negative x, log(x²) exists but log(x) does not. The identity valid for all real x ≠ 0 is log(x²) = 2log|x|. Simplifying an equation must not silently enlarge or shrink its original domain.
08 / Change of base and calculator use
Use the same logarithm base in the numerator and denominator.
loga(N) = ln(N)/ln(a) = log₁₀(N)/log₁₀(a)
To see why, write aˣ = N and take a convenient logarithm: x ln(a) = ln(N). Since a ≠ 1, ln(a) is nonzero and you can divide.
log₇(50) ≈ 2.01038
log₄(N) = ½log₂(N), for N > 0
Unless another base is stated, a calculator’s “log” key normally means base 10 and “ln” means base e. Keep enough digits internally and round only the final requested answer. A rounded decimal is not an exact logarithm.
09 / A logarithmic scale compares ratios
Equal score changes correspond to equal multipliers.
Define a fictional score S = 3log₁₀(q/q₀), where q and the reference q₀ are positive quantities measured in the same units. Their ratio is dimensionless.
q = q₀ ⇒ S = 0
q = 100q₀ ⇒ S = 6
S₂ − S₁ = 3log₁₀(q₂/q₁)
A score increase of 3 means the quantity has multiplied by 10; it does not mean the quantity increased by 3. To recover a ratio from a score difference ΔS, use q₂/q₁ = 10ΔS/3.
10 / Your turn
Turn each logarithm back into a power when unsure.
Use a > 0, a ≠ 1 and positive variable inputs unless a question says otherwise.
02 · Inverse statement
Rewrite log₅(1/25) = −2 using a power.
Hint
Keep the same base.
Worked solution
5⁻² = 1/25.
03 · Exact values
Find log₂(√2), log₇(1) and log1/2(8).
Hint
Write each input as a power of its base.
Worked solution
1/2, 0 and −3.
04 · A product and quotient
Simplify log₂(6) + log₂(8) − log₂(3).
Hint
Combine to log₂(48/3).
Worked solution
log₂(16) = 4.
05 · A power
Simplify 2log₅(10) − log₅(4).
Hint
Combine to log₅(100/4).
Worked solution
log₅(25) = 2.
06 · Expand
Expand loga(x²y³/z), with x,y,z > 0.
Hint
The denominator gives a subtraction.
Worked solution
2loga(x) + 3loga(y) − loga(z).
07 · Condense
Write ½loga(x) − 2loga(y) as one logarithm.
Hint
Move each coefficient to a power first.
Worked solution
loga(√x/y²).
08 · Given one logarithm
If p = logb(27), express logb(9b²) in terms of p.
Hint
logb(3) = p/3.
Worked solution
2p/3 + 2.
09 · Invalid splitting
Is log₁₀(2 + 3) equal to log₁₀(2) + log₁₀(3)?
Hint
The right side is the logarithm of a product.
Worked solution
No. The left is log₁₀(5), and the right is log₁₀(6).
10 · A scale ratio
For S = 5log₁₀(q/q₀), what quantity multiplier gives a score increase of 2?
Hint
2 = 5log₁₀(q₂/q₁).
Worked solution
q₂/q₁ = 102/5 ≈ 2.512.
11 · The domain
At x = −3, which exist as real numbers: ln(x²), 2ln(x), 2ln|x|?
Hint
Check every log input before simplifying.
Worked solution
ln(x²) = ln(9) and 2ln|x| = 2ln(3) exist and are equal. 2ln(x) is undefined as a real expression.
12 · Change of base
Express log₈(32) exactly.
Hint
Use powers of 2, or divide log₂(32) by log₂(8).
Worked solution
5/3.
11 / Recap
Log laws are index laws viewed backwards.
- The logarithm is an exponent; its input is positive.
- Use a positive base other than 1.
- Products add logs, quotients subtract them, powers multiply them.
- A sum inside a logarithm cannot generally be split.
- Keep the original domain when simplifying.
- A logarithmic scale converts quantity ratios to score differences.
Section 1 of 11 · A logarithm is an exponent