To solve an equation with the unknown in a power, either match the bases or take logarithms of both sides. To solve one with the unknown inside a logarithm, combine the logs, convert to an index equation, then reject any answer that makes an argument zero or negative.
This lesson is part of SPM Additional Mathematics indices, surds and logarithms. It relies on the logarithm laws and domain checks.
Which method should I choose?
The form of the equation tells you the method.
| Form | Method | Example |
|---|---|---|
| Both sides are powers of one base | Equate the powers | 4^x = 32 |
| Different bases, one power | Take log of both sides | 3^(2x−1) = 20 |
| Quadratic in a^x | Substitute y = a^x | 2^(2x) − 5(2^x) + 4 = 0 |
| Logs with the same base | Combine, then convert to an index | log₂(x+3) + log₂(x−3) = 4 |
Worked example: unknown in the power
Solve 3^(2x−1) = 20, correct to 3 decimal places.
Take common logs of both sides: (2x − 1) log 3 = log 20.
Then 2x − 1 = log 20 ÷ log 3 = 1.3010 ÷ 0.4771 = 2.7268.
So 2x = 3.7268 and x ≈ 1.863. Check: 2(1.863) − 1 = 2.726, and 3^2.726 ≈ 20.
The tempting shortcut is to write 2x − 1 = 20 ÷ 3. That treats the base as something you can divide out, but the unknown is an exponent, so the log is needed.
Worked example: logs with a domain check
Solve log₂(x + 3) + log₂(x − 3) = 4.
- Combine: log₂[(x + 3)(x − 3)] = 4, so log₂(x² − 9) = 4.
- Convert to an index: x² − 9 = 2⁴ = 16.
- Solve: x² = 25, so x = 5 or x = −5.
- Check the domain. Both x + 3 and x − 3 must be positive, so x > 3.
x = 5 is accepted and x = −5 is rejected, since x − 3 = −8 is negative. Substitute back: log₂ 8 + log₂ 2 = 3 + 1 = 4.
The substitution method
Solve 2^(2x) − 5(2^x) + 4 = 0. Since 2^(2x) = (2^x)², let y = 2^x.
Then y² − 5y + 4 = 0, so (y − 1)(y − 4) = 0 and y = 1 or y = 4.
Convert back: 2^x = 1 gives x = 0, and 2^x = 4 gives x = 2. Both work, because y = 2^x must be positive and both y values are.
The mistake that costs marks
Two slips cost marks here. The first is dividing by the base, as above. The second is forgetting the domain in a log equation, so both x = 5 and x = −5 appear as answers.
A habit that catches both: after finding an answer, substitute it into the original equation and ask whether every logarithm is still defined.
Check yourself
Solve log₃(x + 6) − log₃ x = 2.
Answer
Combine: log₃[(x + 6) ÷ x] = 2, so (x + 6) ÷ x = 3² = 9.
Then x + 6 = 9x, so 8x = 6 and x = 0.75.
Domain: x > 0 and x + 6 > 0, so 0.75 is accepted.
Check: log₃ 6.75 − log₃ 0.75 = log₃ 9 = 2.
What to study next
Try the mixed chapter practice set. If logs and indices are still tangled, return to applying index laws with fractional powers.
For a teacher to work through equation types with you, see online one-to-one Additional Mathematics tuition. The word problem structure worksheet helps when the equation comes from a story.