Laws of logarithms

The three laws

For the same base a:

• log_a x + log_a y = log_a (xy)
• log_a x - log_a y = log_a (x/y)
• k log_a x = log_a (x^k)

Special case: log_a (1/x) = -log_a x.

Common mistakes

log_a (x + y) is not log_a x + log_a y, and (log_a x)/(log_a y) is not log_a (x/y).

Solving log equations

Use the laws to write each side as a single log, then convert to index form. Check each answer: the expression inside every log must be positive.

Worked example

Evaluate 2 log_3 6 - log_3 4.

1. 2 log_3 6 = log_3 36
2. log_3 36 - log_3 4 = log_3 9
3. 3² = 9

Answer: 2

Check yourself

1. Evaluate log_2 12 - log_2 3.
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   2. log_2 (12/3) = log_2 4 = 2.

2. Evaluate log_6 4 + log_6 9.
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   2. log_6 36 = 2.

3. Evaluate 3 log_5 10 - log_5 8.
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   3. log_5 1000 - log_5 8 = log_5 125 = 3.

4. Which is equal to log x + log y?
   • log (x + y)
   • (log x)(log y)
   • log x / log y
   • log (xy)
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   log (xy). Adding logs of the same base gives the log of the product.

5. Write log_a (x³/y) in terms of log_a x and log_a y.
   • (log_a x³)/(log_a y)
   • 3 log_a x - log_a y
   • 3 log_a x + log_a y
   • log_a 3x - log_a y
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   3 log_a x - log_a y. Divide: log_a x³ - log_a y. Then the power: 3 log_a x - log_a y.

6. Solve log_2 x + log_2 (x + 2) = 3.
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   2. log_2 (x(x + 2)) = 3, so x² + 2x = 8 and (x + 4)(x - 2) = 0. x = -4 is rejected because log_2 (-4) is undefined, so x = 2.

7. Solve log_3 (2x + 1) - log_3 (x - 1) = 2. Give your answer as a fraction in its simplest form.
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   10/7. (2x + 1)/(x - 1) = 9, so 2x + 1 = 9x - 9 and 7x = 10. x = 10/7, which makes x - 1 positive.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 2 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.