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TMUA Exponentials and Logarithms

An exponential has the variable in the power, as in axa^x, and a logarithm is its inverse. Expect to use the log laws and to solve equations that reduce to a simpler form.

Paper 1 and Paper 2 topic. Specification codes: MM5.1, MM5.2, MM5.3. Last checked 12 November 2026.

Logarithm laws

  • log⁡ax+log⁡ay=log⁡a(xy)\log_a x+\log_a y=\log_a(xy) and log⁡ax−log⁡ay=log⁡axy\log_a x-\log_a y=\log_a\dfrac xy.
  • log⁡axk=klog⁡ax\log_a x^k=k\log_a x, log⁡aa=1\log_a a=1 and log⁡a1=0\log_a1=0.
  • log⁡ab=c\log_a b=c means ac=ba^c=b.

Solving $a^x=b$

Take logs, or write both sides as powers of the same base. When the equation has a mix of powers, look for a substitution such as u=2xu=2^x.

Check validity

You can only take the log of a positive number, so reject any solution that makes a log argument zero or negative.

Special cases and graphs

Know that log⁡aak=k\log_a a^k=k, so log⁡232=5\log_2 32=5. Negative and fractional answers are fine: log⁡218=−3\log_2\tfrac18=-3 because 2−3=182^{-3}=\tfrac18. The specification excludes change-of-base questions.

The graph of y=axy=a^x with a>1a>1 always passes through (0,1)(0,1), is always positive and never meets the xx-axis. The graph of y=log⁡axy=\log_ax is its reflection in the line y=xy=x.

If a question gives log⁡a2=p\log_a2=p and asks for log⁡a8\log_a8, use the power law: log⁡a8=3log⁡a2=3p\log_a8=3\log_a2=3p. Express every number in the problem through the ones you are given.

Check any answer by substituting it back, especially after a substitution or after squaring, since a spurious solution can appear. A positive base raised to any power stays positive, so 2x=−42^x=-4 has no solution.

Worked examples

Example 1

Solve 2x+1=3⋅2x−82^{x+1}=3\cdot2^x-8.

Show the worked solution

2⋅2x=3⋅2x−82\cdot2^x=3\cdot2^x-8, so 2x=82^x=8 and x=3x=3.

Answer: x=3x=3

Example 2

Solve log⁡2x+log⁡2(x−2)=3\log_2x+\log_2(x-2)=3.

Show the worked solution

log⁡2(x(x−2))=3\log_2\left(x(x-2)\right)=3, so x2−2x=8x^2-2x=8 and (x−4)(x+2)=0(x-4)(x+2)=0. Only x=4x=4 is valid, since x=−2x=-2 makes the logs undefined.

Answer: x=4x=4

Example 3

Solve 32x−4⋅3x+3=03^{2x}-4\cdot3^x+3=0.

Show the worked solution

Let u=3xu=3^x: u2−4u+3=0u^2-4u+3=0, so u=1u=1 or u=3u=3. Then x=0x=0 or x=1x=1.

Answer: x=0x=0 or x=1x=1

Common mistakes

  • Writing log⁡(x+y)=log⁡x+log⁡y\log(x+y)=\log x+\log y.
  • Forgetting to reject a solution that makes a log argument negative.
  • Treating 2x+12^{x+1} as 2x+12^x+1.
  • Solving u2−4u+3=0u^2-4u+3=0 and stopping at uu without converting back to xx.

Test yourself on this topic with our free practice questions.

Quick answers

What are the laws of logarithms?

Adding logs multiplies the arguments, subtracting divides them, and a power in the argument comes out as a multiplier.

How do you solve an exponential equation without a calculator?

Write both sides with the same base, or use a substitution to make a quadratic, so the answer is a whole number or simple fraction.

Why do some log solutions get rejected?

A logarithm is only defined for positive arguments, so any value that makes an argument zero or negative is not a solution.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

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