Indices and logarithms: the laws, with worked examples

Indices and logarithms appear in almost every JAMB, WAEC and NECO Mathematics paper. They look intimidating at first, but the whole topic rests on a handful of laws. Once those laws are automatic, most questions take less than a minute.

Part 1: Indices

An index (plural: indices) tells you how many times a number is multiplied by itself. In 2⁵, 2 is the base and 5 is the index (or power):

2⁵ = 2 × 2 × 2 × 2 × 2 = 32

The laws of indices

These laws only work when the bases are the same. That leads to the most important technique in this topic.

The key technique: make the bases the same

Rewrite every number as a power of the same prime number. For example:

• 4 = 2², 8 = 2³, 16 = 2⁴, 32 = 2⁵
• 9 = 3², 27 = 3³, 81 = 3⁴
• 25 = 5², 125 = 5³

Worked example 1: Simplify

Simplify 8^(2/3) × 16^(−1/4).

• 8^(2/3) = (2³)^(2/3) = 2² = 4
• 16^(−1/4) = (2⁴)^(−1/4) = 2⁻¹ = 1/2
• 4 × 1/2 = 2

Worked example 2: Evaluate a fractional index

Evaluate (81/16)^(−3/4).

A negative index means take the reciprocal: (81/16)^(−3/4) = (16/81)^(3/4).

Fourth root of 16/81 = 2/3. Cube it: (2/3)³ = 8/27.

Tip: with fractional indices, take the root first (the denominator), then the power (the numerator). The numbers stay smaller.

Worked example 3: Solve an index equation

Solve 2^(x+1) = 32.

Write 32 as a power of 2: 32 = 2⁵.

So 2^(x+1) = 2⁵. When the bases are equal, the indices are equal:

x + 1 = 5, so x = 4.

Worked example 4: Index equation with different bases

Solve 9^(x−1) = 27^x.

Write both as powers of 3:

• 9^(x−1) = (3²)^(x−1) = 3^(2x−2)
• 27^x = (3³)^x = 3^(3x)

So 2x − 2 = 3x, which gives x = −2.

Worked example 5: An equation that becomes a quadratic

Solve 2^(2x) − 6(2^x) + 8 = 0.

Let y = 2^x. Then 2^(2x) = y², so the equation becomes:

y² − 6y + 8 = 0 → (y − 2)(y − 4) = 0 → y = 2 or y = 4.

• 2^x = 2 gives x = 1
• 2^x = 4 gives x = 2

So x = 1 or x = 2.

Part 2: Logarithms

A logarithm answers the question: what power do I raise the base to, to get this number?

If aˣ = N, then logₐN = x.

For example, 10² = 100, so log₁₀100 = 2. And 2⁵ = 32, so log₂32 = 5.

Logarithms are the inverse of indices, which is why their laws mirror the laws of indices.

The laws of logarithms

Worked example 6: Using given values

Given that log₁₀2 = 0.3010 and log₁₀3 = 0.4771, find log₁₀12.

Write 12 using 2 and 3: 12 = 2² × 3.

log 12 = log(2² × 3) = 2 log 2 + log 3 = 2(0.3010) + 0.4771 = 0.6020 + 0.4771 = 1.0791.

Worked example 7: Simplify

Simplify log₁₀50 + log₁₀2.

Using the product law: log 50 + log 2 = log(50 × 2) = log 100 = 2.

Worked example 8: Simplify an expression

Simplify (log 8 − log 2) ÷ log 4.

• log 8 − log 2 = log(8 ÷ 2) = log 4
• log 4 ÷ log 4 = 1

Notice that we could not simplify log 4 ÷ log 4 using the quotient law. The quotient law applies to log(M/N), not to (log M) ÷ (log N). In this case the two are simply equal, so the answer is 1.

Worked example 9: Solve a log equation

Solve log₂x + log₂(x − 2) = 3.

Combine using the product law: log₂[x(x − 2)] = 3.

Convert to index form: x(x − 2) = 2³ = 8.

x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2.

Check: the logarithm of a negative number is undefined, so x = −2 is rejected (log₂(−2) is not defined). x = 4.

Always check your answers in the original equation when solving log equations.

Common mistakes

• Adding indices when the bases differ. 2³ × 3² is not 6⁵. It's 8 × 9 = 72.
• Multiplying instead of adding in the multiplication law. 3² × 3⁴ = 3⁶, not 3⁸.
• Thinking a⁰ = 0. Any non-zero number to the power 0 is 1.
• Getting negative indices wrong. 2⁻³ = 1/8, not −8.
• Inventing log laws. log(M + N) is not log M + log N. And (log M)/(log N) is not log(M/N).
• Forgetting to check log equation solutions for values that make a logarithm undefined.

How to practise

1. Learn the laws until you can write them from memory.
2. Practise rewriting numbers as prime powers until it's instant.
3. Do mixed questions (simplify, evaluate and solve) so you recognise each type quickly.
4. Read the worked solution for every question you miss.

Levia's Mathematics levels include indices, surds and logarithms questions at increasing difficulty, each with a step-by-step explanation. Further Mathematics takes them further, with harder index and log equations.