Product and Quotient Properties of Logarithms, precalculus lesson

Two years of doubling followed by three more years of doubling make five years of doubling, because $2^2 \cdot 2^3 = 2^5$ . When powers of the same base are multiplied, their exponents add. Logarithms read off exponents, so they turn products into sums and quotients into differences. Before calculators, this is how long multiplications were done, by adding numbers looked up in tables of logarithms.

Key terms

Logarithm of a product

Definition Let $b > 0$ with $b \ne 1$ , and let $M > 0$ and $N > 0$ . Then

\log_b(MN) = \log_b M + \log_b N.

To prove it, put $u = \log_b M$ and $v = \log_b N$ , so that $b^u = M$ and $b^v = N$ . Then $MN = b^u b^v = b^{u + v}$ , and by the definition of the logarithm $\log_b(MN) = u + v$ .

In plain words The logarithm of a product is the sum of the logarithms. It is the rule “multiply powers, add exponents” read through logarithms.

Example Since $8 \cdot 32 = 256$ ,

\begin{gathered} \log_2(8 \cdot 32) = \log_2 256 = 8 \\[0.4em] \text{and} \\[0.4em] \log_2 8 + \log_2 32 = 3 + 5 = 8. \end{gathered}

The rule also works when the separate logarithms are not nice numbers. The values $\log 2 \approx 0.30103$ and $\log 5 \approx 0.69897$ add up to exactly $\log 10 = 1$ .

Logarithm of a quotient

Definition Let $b > 0$ with $b \ne 1$ , and let $M > 0$ and $N > 0$ . Then

\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N.

With $u = \log_b M$ and $v = \log_b N$ as before,

\frac{M}{N} = \frac{b^u}{b^v} = b^{u - v},

so the logarithm of the quotient is $u - v$ . Taking $M = 1$ gives the special case $\log_b \frac{1}{N} = -\log_b N$ .

In plain words The logarithm of a quotient is the logarithm of the top minus the logarithm of the bottom. It is not the logarithm of the top divided by the logarithm of the bottom.

Example Since $\frac{243}{9} = 27$ ,

\begin{gathered} \log_3\left(\frac{243}{9}\right) = \log_3 27 = 3 \\[0.4em] \text{and} \\[0.4em] \log_3 243 - \log_3 9 = 5 - 2 = 3. \end{gathered}

Dividing the logarithms instead gives $\frac{5}{2}$ , which is wrong.

Positive arguments

Definition The product and quotient rules require each of $M$ and $N$ to be positive, not only their product or quotient. If $M < 0$ and $N < 0$ , then $MN > 0$ , so $\log_b(MN)$ is defined while $\log_b M$ and $\log_b N$ are not. For any real $M$ and $N$ with $MN > 0$ , which is the same as $\frac{M}{N} > 0$ , the correct statements are

\begin{gathered} \log_b(MN) = \log_b \lvert M \rvert + \log_b \lvert N \rvert, \\[0.8em] \log_b\left(\frac{M}{N}\right) = \log_b \lvert M \rvert - \log_b \lvert N \rvert. \end{gathered}

In plain words Before you split a logarithm, check that every piece you write has a positive argument. Splitting can shrink the set of inputs where the expression makes sense.

Example The expression $\log_2\big((-4)(-8)\big)$ is fine and equals $\log_2 32 = 5$ , but $\log_2(-4) + \log_2(-8)$ is undefined. The version with absolute values gives $\log_2 4 + \log_2 8 = 2 + 3 = 5$ . With a variable, the expression $\log\big((x + 1)(x - 3)\big)$ is defined when $x < -1$ or $x > 3$ , while $\log(x + 1) + \log(x - 3)$ is defined only when $x > 3$ . At $x = -2$ the first equals $\log 5$ and the second is undefined.

Common base

Definition The product and quotient rules combine logarithms that have the same base $b$ . A sum $\log_b M + \log_c N$ with $b \ne c$ is in general not equal to $\log_b(MN)$ or to $\log_c(MN)$ . To combine logarithms with different bases, first rewrite them with one base, which the change of base formula in the next lesson does.

In plain words You can only merge logarithms that have the same base. The common logarithm $\log$ and the natural logarithm $\ln$ have different bases, so they do not merge either.

Example The sum

\log_2 8 + \log_4 8 = 3 + \frac{3}{2} = \frac{9}{2},

since $4^{3/2} = 8$ . Merging the arguments gives $64$ , but $\log_2 64 = 6$ and $\log_4 64 = 3$ , and neither is $\frac{9}{2}$ .

Expanding

Definition Expanding a logarithm means using the product and quotient rules to rewrite the logarithm of a product or quotient as a sum and difference of logarithms of its factors. Each step is valid only where every new argument is positive, so an expanded form comes with a condition on the variables.

In plain words Expanding breaks one complicated logarithm into several simple ones. In calculus this makes derivatives of products and quotients much easier, a method called logarithmic differentiation.

Example For $x > 0$ and $y > 0$ ,

\begin{aligned} &\log_5\left(\frac{25x}{y}\right) \\[0.8em] &\quad = \log_5 25 + \log_5 x - \log_5 y \\[0.4em] &\quad = 2 + \log_5 x - \log_5 y. \end{aligned}

For a single variable, consider $\ln \frac{x(x + 1)}{x - 2}$ . When $x > 2$ all three factors are positive, so

\begin{gathered} \begin{aligned} &\ln \frac{x(x + 1)}{x - 2} \\[0.8em] &\quad = \ln x + \ln(x + 1) - \ln(x - 2), \end{aligned} \\[0.4em] x > 2. \end{gathered}

The left side is also defined for $-1 < x < 0$ , where the quotient is positive, but there $\ln x$ and $\ln(x - 2)$ are undefined. The equation holds only for $x > 2$ .

Condensing

Definition Condensing means using the rules from right to left to combine sums and differences of logarithms with the same base into a single logarithm. For $M, N, P > 0$ ,

\begin{aligned} &\log_b M + \log_b N - \log_b P \\[0.4em] &\quad = \log_b \frac{MN}{P}. \end{aligned}

The condensed expression may be defined for more values of the variable than the original one, so keep track of the original domain.

In plain words Condensing packs several logarithms into one, which is the usual first step in solving an equation with logarithms.

Example With numbers,

\log_6 4 + \log_6 9 = \log_6 36 = 2.

With a variable, for $x > 0$

\begin{aligned} &\ln x + \ln(x + 3) - \ln 2 \\[0.4em] &\quad = \ln \frac{x(x + 3)}{2}. \end{aligned}

The left side needs $x > 0$ , while the right side is also defined for $x < -3$ . If an equation leads to a value like $x = -4$ , the condensed form accepts it but the original expression does not, so that value is not a solution.

Sums do not split into logarithms

Definition There is no rule that splits the logarithm of a sum or a difference. For positive $M$ and $N$ ,

\log_b(M + N) = \log_b M + \log_b N

holds exactly when $M + N = MN$ , because the right side equals $\log_b(MN)$ and a logarithm is one-to-one. In the same way $\log_b(M - N)$ is in general not $\log_b M - \log_b N$ . What does work is factoring the sum first, for example

\begin{aligned} &\log_b(M + N) \\[0.4em] &\quad = \log_b\left(M\left(1 + \frac{N}{M}\right)\right) \\[0.8em] &\quad = \log_b M + \log_b\left(1 + \frac{N}{M}\right). \end{aligned}

In plain words Logarithms turn products into sums. They do not turn sums into anything simpler. If you see a sum inside a logarithm, try to factor it.

Example A quick numerical check exposes the mistake. The value of $\log(10 + 10)$ is $\log 20 \approx 1.30103$ , but $\log 10 + \log 10 = 2$ . When the sum factors, the product rule applies to the factors. For $x > 0$

\begin{aligned} \ln\left(x^2 + 5x\right) &= \ln\big(x(x + 5)\big) \\[0.4em] &= \ln x + \ln(x + 5). \end{aligned}

Check yourself

Write $\log_4 2 + \log_4 32$ as a single logarithm and evaluate it.

Show the answer

By the product rule the sum is $\log_4 64$ , and $4^3 = 64$ , so the value is $3$ .

Expand $\log_3 \frac{9x}{x + 4}$ for $x > 0$ .

Show the answer

For $x > 0$ the factors $9$ , $x$ and $x + 4$ are positive, so

\begin{aligned} &\log_3 \frac{9x}{x + 4} \\[0.8em] &\quad = 2 + \log_3 x - \log_3(x + 4). \end{aligned}

The left side is also defined for $x < -4$ , but the expanded form is not.

Solve $\log_2 x + \log_2(x - 2) = 3$ .

Show the answer

Condensing gives $\log_2\big(x(x - 2)\big) = 3$ , so $x(x - 2) = 8$ , which means $x^2 - 2x - 8 = 0$ and $x = 4$ or $x = -2$ . The original equation needs $x > 2$ , so $x = -2$ is rejected. The only solution is $x = 4$ , and indeed $\log_2 4 + \log_2 2 = 2 + 1 = 3$ .

Product and Quotient Properties of Logarithms

Key terms

Logarithm of a product

Logarithm of a quotient

Positive arguments

Common base

Expanding

Condensing

Sums do not split into logarithms

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus