Key Takeaways
Over the reals, log₂8+log₂4=3+2=5. For log₂(x−1)+log₂(x+1)=3, retain x>1. Simplification gives x²=9, but only x=3 is valid; a positive product after squaring does not make the original negative arguments at −3 acceptable.
- Adding same-base logarithms corresponds to multiplying their arguments: log₂8 + log₂4 = 5, not the logarithm of 8 + 4.Read the supporting section: Explain the notation through powers
- Each original argument of a real logarithm must be positive; a positive product is not a substitute for that condition.Read the supporting section: A positive product does not replace positive arguments
- After retaining x > 1 in the worked equation, accept only x = 3 from the candidates for x² = 9 and reject x = −3.Read the supporting section: Two candidates, one solution to the original
In this article
Explain the notation through powers
OpenStax’s logarithm definition relates log_b(a)=c to b^c=a, with b>0, b≠1 and a>0 for real logarithms. This article stays within that setting, not complex logarithms.
In the original numerical example, 2³=8 and 2²=4, so the logarithms are 3 and 2. Adding exponents corresponds to 2³×2²=2⁵, which explains log₂(8×4)=5 rather than merely asking you to memorize a switch from addition to multiplication.
A small numerical check rejects the wrong rule
log₂8+log₂4 = log₂(8+4) = log₂12
log₂8+log₂4 = 3+2 = 5 = log₂32
A positive product does not replace positive arguments
OpenStax’s logarithmic properties document the same-base product rule. Before applying it to log₂(x−1)+log₂(x+1), retain both x−1>0 and x+1>0. Their intersection is x>1.
The standalone expression log₂(x²−1) also permits x<−1, but there both original arguments are negative. An expanded domain in a replacement expression cannot overwrite the original conditions. That is how the candidate −3 can slip into a wrong answer.
| Expression | Requirement | Range |
|---|---|---|
| log₂(x−1)+log₂(x+1) | x−1>0 and x+1>0 | x>1 |
| log₂(x²−1) on its own | x²−1>0 | x<−1 or x>1 |
| Combined expression used to solve the original | Retain the original conditions | Still x>1 |
Two candidates, one solution to the original
Solve log₂(x−1)+log₂(x+1)=3. First state x>1. Under that condition, combine to log₂[(x−1)(x+1)]=3 and convert to (x−1)(x+1)=2³=8. Hence x²−1=8, x²=9 and the candidates are 3 and −3.
The value 3 satisfies x>1 and gives log₂2+log₂4=1+2=3. At −3, the original arguments are −4 and −2, invalid for real logarithms. Multiplying those negatives to obtain 8 is not substitution into the original. The solution set is {3}.
The radical-equation candidate check also returns to an original equation, but the error here is expansion of a logarithmic domain, not another name for a squaring step.
Explain every transition
- Base 2 is valid; both original arguments positive give x>1.
- Combine the sum into a product; convert the right side to 2³=8.
- Treat ±3 as candidates, not the final solution set.
- Substitute into the original, retain 3 and explain the rejection of −3.
Change the base without losing the definition
Original variation 1: evaluate log_(1/2)(1/8). Since (1/2)³=1/8, the answer is 3. A base between zero and one is valid; do not silently replace it with 2 or combine it with base-2 logarithms using the same-base rule.
Original variation 2: log₃(x−2)+log₃(x+2)=2. The domain is x>2. Combining gives x²−4=3²=9, with candidates ±√13. Only √13 satisfies the domain. Its two positive original arguments multiply to 9, so the sum of their logarithms is 2.
Original variation 3: log₁8 has no value under this real-logarithm definition. Every real power of 1 is 1, not 8. The negative-exponent and parentheses exercise similarly keeps the operation’s object explicit rather than changing it because symbols look alike.
| Exercise | Answer | Essential check |
|---|---|---|
| log_(1/2)(1/8) | 3 | Valid 0<b<1; reconstruct the power |
| log₃(x−2)+log₃(x+2)=2 | x=√13 | x>2; reject −√13 |
| log₁8 | Undefined in this real-logarithm setting | The base cannot be 1 |
Record the missed condition, not just the final number
If you added arguments, write: “Use the product rule only with the same valid base and individually positive arguments,” then verify with 8 and 4. If you accepted −3, write: “I checked a positive product but omitted positivity of each original argument.” These errors require different corrections.
Completion means explaining the rule through powers, stating the original domain and accepting or rejecting every candidate with a reason. A correct final number alone does not establish sound reasoning.
