Guides / Logarithms

Logarithm rules: what you can split apart, and what you can't

There are three log rules, and each one works on exactly one shape: a product, a quotient, or a power inside the log. Most lost points come from using a rule on a shape it doesn't cover.

Why it works

A logarithm is an exponent. log₂ 8 = 3 because 2³ = 8. So the log rules are the exponent rules, read backwards:

• b^m · bⁿ = b^m+n: multiplying numbers adds their exponents, so the log of a product is the sum of the logs.
• b^m / bⁿ = b^m−n: dividing subtracts exponents, so the log of a quotient is the difference.
• (b^m)^p = b^mp: raising to a power multiplies the exponent, so the power comes out front as a multiplier.

There is no exponent rule for b^m + bⁿ, and that's why there is no log rule for a sum. Numbers make it obvious: log(2 + 8) = log 10 = 1, but log 2 + log 8 = log 16 ≈ 1.204. (Here and below, log with no base means base 10.)

Example 1: expanding

Expand log₂(8x³/y), for x, y > 0.

1. log₂(8x³) − log₂ yQuotient rule first: the whole numerator minus the denominator.
2. log₂ 8 + log₂(x³) − log₂ yProduct rule on 8 · x³.
3. 3 + 3 log₂ x − log₂ ylog₂ 8 = 3 because 2³ = 8. Power rule on x³: the 3 applies only to x, so it comes out in front of log₂ x only.

Example 2: condensing

Write 2 ln x + ln(x + 1) − ln 5 as a single logarithm, for x > 0.

1. ln(x²) + ln(x + 1) − ln 5Power rule backwards: move every coefficient up into an exponent before combining. The 2 belongs to x only.
2. ln(x²(x + 1)) − ln 5Sum of logs becomes the log of a product. The x + 1 stays in parentheses as one factor.
3. ln x²(x + 1)5Difference of logs becomes the log of a quotient. Whatever is subtracted goes in the denominator.

Notice that ln(x + 1) did not get split into ln x + ln 1. It was a log of a sum, so it stays whole.

Example 3: solving, and the answer that has to go

Solve log x + log(x − 3) = 1.

1. log(x(x − 3)) = 1Product rule, combining the two logs into one.
2. x(x − 3) = 10¹log y = 1 means 10¹ = y. Rewrite in exponential form.
3. x² − 3x − 10 = 0Distribute and move the 10 over.
4. (x − 5)(x + 2) = 0Factor: −5 and 2 multiply to −10 and add to −3.
5. x = 5 or x = −2Two candidates.
6. x = 5Check in the original. x = −2 makes log x into log(−2), which isn't defined, so −2 is rejected. x = 5: log 5 + log 2 = log 10 = 1. It works.

The rejected answer isn't an arithmetic error. Combining the logs on line 1 makes an equation that allows more x values than the original did: x(x − 3) is positive at x = −2, even though x and x − 3 separately are not. The check catches it.

Common mistakes, and the exact line they happen on

1. Splitting the log of a sum

1. log(x + 10) = log x + log 10 = log x + 1This is the mistake. The product rule needs a product inside the log. log x + 1 is log(10x), not log(x + 10). At x = 10: log 20 ≈ 1.301, but log 10 + 1 = 2.

This is the log version of writing (a + b)² = a² + b²: an operation that doesn't spread over addition being spread over addition anyway.

2. Dividing logs as if they were a quotient inside a log

1. log 100log 10 = log 10010This is the mistake. The left side is 2/1 = 2. The right side is log 10 = 1. The quotient rule is about a fraction inside one log, not one log divided by another.

A log divided by a log is what the change-of-base formula produces: log_b M = log M / log b. So log 100 / log 10 = log₁₀ 100 = 2, which is the true value.

3. Using the power rule on a log that's squared

1. (log x)² = 2 log xThis is the mistake. The power rule applies to log(x²), where the square is inside the log. (log x)² squares the log's output. At x = 1000: (log 1000)² = 3² = 9, but 2 log 1000 = 6.

Some textbooks write log² x for (log x)², which makes this even easier to miss. If the exponent is outside the log's parentheses, the power rule doesn't apply.

4. Moving a coefficient to the wrong place

1. 3 log x + log 2Condense to one log.
2. log(3x) + log 2This is the mistake. A coefficient becomes an exponent, not a multiplier: 3 log x = log(x³). The correct condensed form is log(2x³), not log(6x).

5. Keeping an answer that's outside the domain

In Example 3, the last line of algebra gives x = 5 or x = −2. Writing both as the final answer loses the point. Every value has to make every log in the original equation take a positive input. Check them in the first line you were given, not in the combined version, because the combined version is exactly the line that let the bad answer in.

Practice

1. Expand log₃(9x²/y⁴), for x, y > 0.

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   log₃ 9 + 2 log₃ x − 4 log₃ y = 2 + 2 log₃ x − 4 log₃ y.

2. Write 3 log x − log y + log 2 as a single logarithm.

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   Power rule first: log(x³) − log y + log 2. Then combine: log(2x³/y).

3. True or false: log(x + 10) = log x + 1.

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   False. log x + 1 = log x + log 10 = log(10x). At x = 10: log 20 ≈ 1.301, but log 10 + 1 = 2.

4. Solve log₃(x + 1) − log₃(x − 1) = 1.

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   Quotient rule: log₃((x + 1)/(x − 1)) = 1, so (x + 1)/(x − 1) = 3. Then x + 1 = 3x − 3, x = 2. Check: log₃ 3 − log₃ 1 = 1 − 0 = 1 ✓.

5. Solve ln x + ln(x + 2) = ln 15.

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   ln(x(x + 2)) = ln 15, so x² + 2x − 15 = 0, (x + 5)(x − 3) = 0. x = −5 makes ln x undefined, so reject it. x = 3: ln 3 + ln 5 = ln 15 ✓.