A logarithm is only defined when the number inside it is positive. So when you solve a log equation, every answer must be tested: a root that makes any original log argument zero or negative must be rejected.
This lesson builds on combining logarithms and converting between index and log form. It is one of the ways exponential and logarithmic reasoning tests whether you think before you write the final answer.
What is the plan?
- Write the domain first. For each log, require the number inside to be greater than 0, and keep the overlap of all conditions.
- Combine the logs into one log on each side, or one log and a plain number.
- Convert to an ordinary equation using aˣ = y ⇔ x = loga y, or by equating the numbers inside logs with the same base.
- Solve the new equation. Quadratics are common, so the quadratic structure explorer can help you check roots and factors.
- Compare each root with the domain and reject any that fail.
- Substitute your accepted root into the original equation as a final check.
Worked example
Solve log₂ (x + 6) + log₂ x = 4.
Step 1, domain: x + 6 > 0 and x > 0, so the combined condition is x > 0.
Step 2, combine: log₂ [x(x + 6)] = 4.
Step 3, convert: x(x + 6) = 2⁴ = 16.
Step 4, rearrange and solve: x² + 6x − 16 = 0, which factorises as (x + 8)(x − 2) = 0. So x = −8 or x = 2.
Step 5, apply the domain: x = −8 fails because x must be greater than 0. The value x = 2 satisfies it.
Check: log₂ (2 + 6) + log₂ 2 = log₂ 8 + log₂ 2 = 3 + 1 = 4. ✓
Answer: x = 2 only.
Notice that x = −8 would make both x + 6 = −2 and x negative. Neither log could exist, so the root is not just unhelpful but impossible.
The mistake to watch for
The usual slip is to give both roots of the quadratic as the answer.
Mistaken answer: x = 2 or x = −8.
The algebra is correct, but x = −8 is not a solution of the original equation.
The correction is to write the domain condition at the start, then state “x = −8 rejected because x > 0”. Examiners look for that evidence of a decision, so a bare “x = 2” with no mention of the other root may not show enough reasoning. A line of explanation costs a few seconds.
A second, quieter slip is to reject a root for the wrong reason. A root can make one log fail while another log stays fine, so test every log in the original equation, not just the first.
Check yourself
1. Solve log₃ x + log₃ (x − 2) = 1.
Show answer
Domain: x > 0 and x − 2 > 0, so x > 2. Combine: log₃ [x(x − 2)] = 1, so x(x − 2) = 3. Then x² − 2x − 3 = 0, so (x − 3)(x + 1) = 0, giving x = 3 or x = −1. Only x = 3 is greater than 2.
Check: log₃ 3 + log₃ 1 = 1 + 0 = 1. So x = 3.
2. Solve log₂ (3x + 1) = 2 log₂ x, giving your answer to 3 significant figures.
Show answer
Domain: x > 0 (and 3x + 1 > 0, which follows). Rewrite the right side as log₂ x². Equate: 3x + 1 = x², so x² − 3x − 1 = 0. The formula gives x = (3 ± √13)/2, which is about 3.303 or −0.303.
The negative root fails the domain, because log₂ x needs x > 0, even though 3x + 1 would still be positive. So x = 3.30.
3. State the values of x for which log₂ (5 − x) + log₂ (x + 1) is defined.
Show answer
Need 5 − x > 0, so x < 5, and x + 1 > 0, so x > −1. Both must hold, so −1 < x < 5.
Where this leads next
Logs also help when a curve is turned into a straight line. Try linearising an exponential relationship next, then test yourself with the mixed practice set.
If you keep finding a wrong extra root at the end of your solutions, a teacher can look at your written working in online one-to-one Additional Mathematics tuition and build a checking habit that fits you.