Properties of Logarithms: Product, Quotient and Power Rules

Quick answer

Logarithms turn multiplication into addition: log_b(MN) = log_b M + log_b N. Division becomes subtraction, log_b(M/N) = log_b M − log_b N, and a power comes down as a factor, log_b(Mᵏ) = k log_b M. Each rule is an exponent rule read backward, because a logarithm is an exponent. The rules expand one logarithm into several or condense several into one, and they give the change-of-base formula, log_b x = ln x / ln b.

What you'll learn

State and apply the product, quotient and power rules
Derive each rule from the matching exponent rule
Expand and condense logarithmic expressions
Evaluate any logarithm with the change-of-base formula

Three rules

For a base $b > 0$ , $b \ne 1$ , and positive numbers $M$ and $N$ :

Rule	Logarithm form	Exponent rule behind it
Product	$\log_b(MN) = \log_b M + \log_b N$	$b^m \cdot b^n = b^{m+n}$
Quotient	$\log_b \tfrac{M}{N} = \log_b M - \log_b N$	$\tfrac{b^m}{b^n} = b^{m-n}$
Power	$\log_b\left(M^k\right) = k\log_b M$	$\left(b^m\right)^k = b^{mk}$

A check with numbers: $\log_2 32 = 5$ , and $\log_2 4 + \log_2 8 = 2 + 3 = 5$ , since $4 \cdot 8 = 32$ .

Why every log rule is an exponent rule

Name the logarithms: let $m = \log_b M$ and $n = \log_b N$ , so that $M = b^m$ and $N = b^n$ . Then

MN = b^m \cdot b^n = b^{m+n} \quad\Longrightarrow\quad \log_b(MN) = m + n = \log_b M + \log_b N

The exponents add when the powers multiply, and the logarithms are those exponents. The quotient and power rules come out the same way. Every log rule is an exponent rule, because a logarithm is an exponent. Before calculators, this is how people multiplied long numbers: look up two logarithms in a table, add them, and look the sum back up.

Condensing

The rules run backward too. Coefficients go back up as exponents, sums become products, and differences become quotients:

2\ln x + \ln(x + 1) - \ln 3 = \ln\frac{x^2(x + 1)}{3}

Condensing is the key step in solving an equation with several logarithms: one logarithm can be undone, several cannot.

Change of base

Calculators have $\log$ (base $10$ ) and $\ln$ (base $e$ ). For any other base, let $y = \log_b x$ , so $b^y = x$ . Take $\ln$ of both sides and use the power rule:

y\ln b = \ln x \quad\Longrightarrow\quad \log_b x = \frac{\ln x}{\ln b}

Worked examples

Common mistakes

Practice problems

Expand $\log_3 (9x)$ .

Answer

$2 + \log_3 x$

Full solution

$\log_3 9 + \log_3 x$ , and $\log_3 9 = 2$ .
Expand $\displaystyle\ln \frac{x^2 y}{z}$ .

Answer

$2\ln x + \ln y - \ln z$

Full solution

The quotient rule separates $z$ , the product rule separates $x^2$ and $y$ , and the power rule brings the $2$ down.
Expand $\displaystyle\log \frac{\sqrt{x}}{10y}$ .

Answer

$\tfrac{1}{2}\log x - 1 - \log y$

Full solution

$\log x^{1/2} - \log(10y) = \tfrac{1}{2}\log x - (\log 10 + \log y)$ , and $\log 10 = 1$ .
Condense $\log x + \log 5 - \log 2$ .

Answer

$\log \tfrac{5x}{2}$

Full solution

The sum becomes the product $5x$ , and the difference divides by $2$ .
Condense $3\ln 2 - \ln 4$ and simplify.

Answer

$\ln 2$

Full solution

$\ln 2^3 - \ln 4 = \ln \tfrac{8}{4} = \ln 2$ .
Condense $\tfrac{1}{2}\log x + 2\log y$ .

Answer

$\log\left(\sqrt{x}\,y^2\right)$

Full solution

The coefficients become exponents, $x^{1/2}$ and $y^2$ , and the sum becomes a product.
Given $\log 2 \approx 0.3010$ and $\log 3 \approx 0.4771$ , find $\log 18$ .

Answer

About $1.2552$

Full solution

$18 = 2 \cdot 3^2$ , so $\log 18 = \log 2 + 2\log 3 \approx 0.3010 + 0.9542$ .
Find $\log_2 10$ with the change-of-base formula.

Answer

About $3.322$

Full solution

$\tfrac{\ln 10}{\ln 2} \approx \tfrac{2.3026}{0.6931}$ . It lies between $3$ and $4$ , since $2^3 = 8$ and $2^4 = 16$ .
Find $\log_4 8$ exactly.

Answer

$\tfrac{3}{2}$

Full solution

Change to base $2$ : $\tfrac{\log_2 8}{\log_2 4} = \tfrac{3}{2}$ . Check: $4^{3/2} = \left(\sqrt{4}\right)^3 = 8$ .
A student writes $\log(x + 3) = \log x + \log 3$ . What went wrong?

Hint

Test the student’s claim with $x = 1$ .

Answer

There is no rule for the logarithm of a sum. $\log x + \log 3$ equals $\log(3x)$ , not $\log(x + 3)$ .

Full solution

With $x = 1$ : $\log 4 \approx 0.602$ , but $\log 1 + \log 3 \approx 0.477$ . The product rule turns a product inside the logarithm into a sum outside it, never a sum into a sum.

Frequently asked questions

What are the three properties of logarithms?

Product: log_b(MN) = log_b M + log_b N. Quotient: log_b(M/N) = log_b M − log_b N. Power: log_b(Mᵏ) = k log_b M.

Why do the log rules work?

A logarithm is an exponent, so each log rule is an exponent rule in disguise. Multiplying powers adds exponents, so the log of a product is the sum of the logs.

Is log(x + y) equal to log x + log y?

No. There is no rule for the log of a sum. log x + log y equals log(xy).

What is the change-of-base formula?

log_b x = log_c x / log_c b for any base c. With natural logs, log_b x = ln x / ln b.

What does it mean to condense a logarithmic expression?

To combine several logarithms into one, using the rules backward: 2 ln x − ln y = ln(x²/y).

What to learn next

Solving Exponential and Logarithmic EquationsSolve equations with the unknown in an exponent or inside a logarithm: matching bases, taking logs of both sides, condensing logs, and rejecting extraneous solutions.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSF.LE.A.4Linear, Quadratic, and Exponential ModelsFor exponential models, express as a logarithm the solution to ab<sup>ct</sup> = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology.
CCSS.MATH.CONTENT.HSF.BF.B.5Building Functions(+) Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents.

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