MA001: College Algebra

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: $\frac{x^a}{x^b}=x^{a−b}$. The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.

Given any real number $x$ and positive real numbers $M, N$, and $b$, where $b \neq 1$, we will show

$log_b(\frac{M}{N})=log_b(M)−log_b(N)$.

Let $m=log_bM$ and $n=log_bN$. In exponential form, these equations are $b^m=M$ and $b^n=N$. It follows that

$\begin{aligned} \log _{b}\left(\frac{M}{N}\right) &=\log _{b}\left(\frac{b^{m}}{b^{n}}\right) & & \text { Substitute for } M \text { and } N . \\ &=\log _{b}\left(b^{m-n}\right) & & \text { Apply the quotient rule for exponents. } \\ &=m-n & & \text { Apply the inverse property of logs. } \\ &=\log _{b}(M)-\log _{b}(N) & & \text { Substitute for } m \text { and } n . \end{aligned}$

For example, to expand $log \left(\frac{2x^2+6x}{3x+9}\right)$, we must first express the quotient in lowest terms. Factoring and canceling we get,

$\begin{aligned} log \left(\frac{2x^2+6x}{3x+9}\right) &= log \left(\frac{2x(x+3)}{3(x+3)} \right) & & \text { Factor the numerator and denominator. } \\ &=log \left( \frac{2x}{3} \right) & & \text { Cancel the common factors. } \end{aligned}$

$\begin{aligned} log (\frac{2x}{3}) &= log(2x)−log(3) \\ &=log(2)+log(x)−log(3) \end{aligned}$

THE QUOTIENT RULE FOR LOGARITHMS

The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.

$log_b(\frac{M}{N})=log_bM−log_bN$

HOW TO

Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.

1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

EXAMPLE 2

Using the Quotient Rule for Logarithms

Expand $log_2 \left(\frac{15x(x−1)}{(3x+4)(2−x)}\right)$.

Solution

$\log _{2}\left(\frac{15 x(x-1)}{(3 x+4)(2-x)}\right)=\log _{2}(15 x(x-1))-\log _{2}((3 x+4)(2-x))$

Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.

$\begin{array}{ll} log_2(15x(x−1))−log_2((3x+4)(2−x)) &=[log_2(3)+log_2(5)+log_2(x)+log_2(x−1)]−[log_2(3x+4)+log_2(2−x)] \\ &=log_2(3)+log_2(5)+log_2(x)+log_2(x−1)−log_2(3x+4)−log_2(2−x) \end{array}$

Analysis

There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for $x=−\frac{4}{3}$ and $x=2$. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that $x > 0, x > 1, x > − \frac{4}{3}$, and $x < 2$. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

TRY IT #2

Expand $log_3 \left(\frac{7x^2+21x}{7x(x−1)(x−2)}\right)$.