Convert between exponential and logarithmic form, and apply log rules to simplify and solve — a logarithm is just an exponent asking a question in a different order. Every question comes with a written walkthrough of the rule applied.
The product rule applies to log(a×b), not log(a+b) — there's no rule that simplifies the log of a sum at all.
log_b(1) = 0 and log_b(b) = 1 for any valid base b — these two facts get swapped under time pressure more than almost any other rule in this subtopic.
In (log a)ⁿ, the exponent applies to the whole logarithm, not to a — that's a completely different expression from log(aⁿ), where the power rule actually applies.
Straight from the bank — one per difficulty tier. Reveal the answer to see the explanation you'd get in a real session.
Write 3⁴ = 81 in logarithmic form.
A — log₃(81) = 4. The base of the exponential (3) becomes the base of the log, the exponent (4) becomes the log's result, and 81 stays what's being evaluated: log₃(81) = 4. log₄(81) = 3 (B) swaps which number becomes the base and which becomes the result.
Solve for x: log₂(x) + log₂(5) = log₂(40)
A — 8. Combine the left side using the product rule: log₂(x) + log₂(5) = log₂(5x). Since log₂(5x) = log₂(40), the arguments must be equal: 5x = 40, so x = 8. 35 (B) comes from treating the logs as if they could be dropped and the arguments simply subtracted — computing 40 − 5 instead of dividing 40 by 5.
Solve for x: 5^(2x+1) = 15
A — x ≈ 0.34. Take the log of both sides: (2x+1)log 5 = log 15, so 2x+1 = log15/log5 ≈ 1.683. Solving, 2x ≈ 0.683, so x ≈ 0.34. x ≈ 0.68 (B) comes from correctly reaching 2x ≈ 0.683 but forgetting the final division by 2 — reporting that intermediate value as x directly.
Logarithms feel abstract until the exponential-form conversion becomes automatic — after that, most problems are just the product, quotient, and power rules in different combinations.
Before applying any rule, convert between exponential and logarithmic form if the problem doesn't already give you the form you want. Untimed practice is where that conversion becomes reflexive.
Move to timed sessions once rule selection is fast. Equations that require taking the log of both sides run a bit longer than direct evaluation.
Pair logarithms with exponents in a mock — they're the same relationship read in two directions.
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