Exponentials and logarithms: 6.5 Solve equations of the form a^x = b.

The idea

An equation aˣ = b has the unknown in the exponent, so ordinary rearrangement cannot isolate x. Take logarithms of both sides: log(aˣ) = log b, then use the power law to bring x down, giving x log a = log b and hence x = log b ÷ log a. Any consistent base works; choosing base 10 or base e suits most calculators. For example, 3ˣ = 20 gives x = log 20 ÷ log 3 ≈ 1.3010 ÷ 0.4771 ≈ 2.73. Check by substitution: 3^2.73 ≈ 20. If b is a recognisable power of a, equate indices directly, as with 2ˣ = 32 giving x = 5. Equations such as 5^(2x−1) = 40 are solved the same way, keeping the whole exponent intact before dividing.

What earns the marks

• Takes logarithms of both sides of aˣ = b using a consistent base, for example log₁₀ or ln.
• Applies the power law log(aˣ) = x log a to bring the exponent down as a product.
• Rearranges to x = log b ÷ log a and evaluates to a sensible number of significant figures.
• Recognises cases where b is an exact power of a and equates indices instead, for example 2ˣ = 32 giving x = 5.
• Handles a compound exponent such as 5^(2x−1) = 40 by keeping the exponent intact, then solving the resulting linear equation.
• Checks the solution by substituting back into the original equation and comparing values.

Common mistakes and how to put them right

• Dividing b by a instead of taking logarithms.

  Put it right by taking logs of both sides first.

• Writing log b ÷ log a as log(b ÷ a).

  Put it right by keeping the two logarithms separate.

• Dropping the exponent when the base is a power of the other side.

  Put it right by equating indices only when the bases match exactly.

Exam tips

• State the logarithm step explicitly before using the power law, so the method is visible.
• Give the final answer to three significant figures unless the question specifies otherwise.
• Substitute the answer back into the original equation as a quick numerical check.