Audited 05 Aug 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Condense Logarithms Calculator

Condense log A + log B − log C into a single logarithm, log(AB/C), using the product and quotient rules. Works for any base shared by all three terms.

Condense Logarithms Calculator

Argument of the condensed logarithm
12
The single argument AB/C, a pure number; log A + log B − log C equals the logarithm of this value in the same base.

Background.

Logarithms were invented to turn multiplication into addition. After John Napier published his tables in 1614, astronomers could multiply two long numbers by looking up their logarithms, adding them, and reading the product back from the table, and every slide rule built afterwards did the same with sliding scales. The rule behind that trick, log A + log B = log(AB), is the one at work when you condense an expression, together with its partner for division, log A − log B = log(A/B).

Condensing runs those rules in the opposite direction to expanding. Instead of splitting one logarithm into several, you gather a sum and difference of logarithms into a single one, and that is usually the step that unlocks a logarithmic equation: once a side is one log, it can be rewritten in exponential form and solved.

The page handles the most common shape, log A + log B − log C, and returns the single argument AB/C. All three logarithms must share a base, but which base does not matter — the condensed argument is the same for base 10, base e or base 2. The arguments must be positive, because a real logarithm of zero or a negative number does not exist. Coefficients other than 1 need the power rule first, and expressions containing variables have to be condensed by hand; the questions below show how to fit both kinds into the three fields.

What is condense logarithms calculator?

Condensing logarithms means rewriting a sum or difference of logarithms with the same base as one logarithm. The product rule turns added logarithms into the logarithm of a product, and the quotient rule turns a subtracted logarithm into division, so log_b A + log_b B − log_b C = log_b(AB/C). The output here is that combined argument, AB/C, a pure number; its logarithm in your chosen base equals the original expression.

How to use this calculator.

  1. Enter the arguments of the two added logarithms in the first and second fields. If only one logarithm is added, put 1 in the second field — log 1 is zero, so it changes nothing.
  2. Enter the argument of the subtracted logarithm in the third field, or 1 if nothing is subtracted.
  3. Apply the power rule before entering any term with a coefficient: 2 log 3 becomes log 9, and ½ log 16 becomes log 4.
  4. Read the result as the argument of one logarithm in the same base as the original terms, so log 8 + log 9 − log 6 condenses to log 12.

The formula.

log(A)+log(B)-log(C)=log(AB/C)

Two identities do the work. The product rule, log_b A + log_b B = log_b(AB), holds because logarithms are exponents: if b^x = A and b^y = B, then b^(x+y) = AB. The quotient rule follows the same way — with b^z = C, b^(x+y−z) = AB/C — so log_b A + log_b B − log_b C = log_b(AB/C). The page computes AB/C directly, multiplying the first two arguments and dividing by the third. The base b never enters that arithmetic, which is why a single answer serves every base. The result grows in proportion to A and to B and shrinks in proportion to C, so doubling C halves it. Each argument must be positive, and the fields accept values from 0.000001 to 1,000,000,000.

A worked example.

Example

Condense log 8 + log 9 − log 6. The two added logarithms combine by the product rule into log(8 × 9) = log 72, and subtracting log 6 divides that argument: 72 ÷ 6 = 12. The expression is therefore log 12, and the page returns 12. A check in base 10: log 8 ≈ 0.90309, log 9 ≈ 0.95424 and log 6 ≈ 0.77815, so 0.90309 + 0.95424 − 0.77815 = 1.07918, which is log 12 to five places. In base 2 the same expression is 3 + log₂ 9 − log₂ 6, about 3.585, which is log₂ 12. The tempting slip is to add the arguments instead: log 8 + log 9 is log 72, not log 17.

argument A8
argument B9
argument C6

Frequently asked questions.

Why is log A + log B not the same as log(A + B)?
Because logarithms turn multiplication, not addition, into addition. In base 10, log 8 + log 9 is about 1.8573, which is log 72, while log 17 is only about 1.2304. There is no rule that simplifies the logarithm of a sum, so an expression like log(A + B) cannot be split or condensed any further.
How do I condense a term with a coefficient, such as 2 log 3?
Move the coefficient inside with the power rule, n log_b A = log_b(Aⁿ), before using the fields. 2 log 3 becomes log 9 and 3 log 2 becomes log 8. So 2 log 3 + log 8 − log 6 is entered as 9, 8 and 6, and it condenses to log 12, exactly like the worked example.
What if two logarithms are being subtracted?
Combine the subtracted terms first: log A + log B − log C − log D equals log(AB / (CD)), since subtracting two logs is the same as subtracting the log of their product. Enter C × D in the third field. For instance, log 8 + log 9 − log 2 − log 3 uses 6 there and condenses to log 12.
Does it matter whether the logs are base 10, natural or base 2?
Not for the condensed argument, provided every term uses the same base. The rules hold in any base, so log 8 + log 9 − log 6 condenses to log 12 and ln 8 + ln 9 − ln 6 to ln 12. Terms in different bases must first be converted with the change-of-base formula.
Why does every argument have to be positive?
No real power of a positive base produces zero or a negative number, so real logarithms exist only for positive arguments. Something like log(−4) has no real value and cannot be combined with anything. The same fact is why, when solving equations, any solution that makes an original argument zero or negative must be thrown out.
How does condensing help solve a logarithmic equation?
It reduces several logs to one, which can then be rewritten in exponential form. For log x + log(x − 3) = 1 in base 10, condensing gives log(x(x − 3)) = 1, so x(x − 3) = 10 and x = 5 or x = −2. Only x = 5 keeps both original arguments positive, so it is the single solution.

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log(A)+log(B)-log(C)=log(AB/C)
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