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: [latex]{x}^{\frac{a}{b}}={x}^{a-b}[/latex]. 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 xand positive real numbers M, N, and b, where [latex]b\ne 1[/latex], we will show
[latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{=}{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)[/latex].
Let [latex]m={\mathrm{log}}_{b}M[/latex] and [latex]n={\mathrm{log}}_{b}N[/latex]. In exponential form, these equations are [latex]{b}^{m}=M[/latex] and [latex]{b}^{n}=N[/latex]. It follows that
[latex]\begin{cases}{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\hfill & ={\mathrm{log}}_{b}\left(\frac{{b}^{m}}{{b}^{n}}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m-n}\right)\hfill & \text{Apply the quotient rule for exponents}.\hfill \\ \hfill & =m-n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{cases}[/latex]
For example, to expand [latex]\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right)[/latex], we must first express the quotient in lowest terms. Factoring and canceling we get,
[latex]\begin{cases}\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right) & =\mathrm{log}\left(\frac{2x\left(x+3\right)}{3\left(x+3\right)}\right)\hfill & \text{Factor the numerator and denominator}.\hfill \\ & \text{ }=\mathrm{log}\left(\frac{2x}{3}\right)\hfill & \text{Cancel the common factors}.\hfill \end{cases}[/latex]
Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.
[latex]\begin{cases}\mathrm{log}\left(\frac{2x}{3}\right)=\mathrm{log}\left(2x\right)-\mathrm{log}\left(3\right)\hfill \\ \text{ }=\mathrm{log}\left(2\right)+\mathrm{log}\left(x\right)-\mathrm{log}\left(3\right)\hfill \end{cases}[/latex]
A General Note: 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.
[latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N[/latex]
How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.
- Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
- Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
- 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 [latex]{\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)[/latex].
Solution
First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.
[latex]{\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)={\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right)[/latex]
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.
[latex]\begin{cases}{\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right) \\\text{ }= \left[{\mathrm{log}}_{2}\left(3\right)+{\mathrm{log}}_{2}\left(5\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)\right]-\left[{\mathrm{log}}_{2}\left(3x+4\right)+{\mathrm{log}}_{2}\left(2-x\right)\right]\hfill \\ \text{ }={\mathrm{log}}_{2}\left(3\right)+{\mathrm{log}}_{2}\left(5\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)-{\mathrm{log}}_{2}\left(3x+4\right)-{\mathrm{log}}_{2}\left(2-x\right)\hfill \end{cases}[/latex]
Analysis of the Solution
There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\frac{4}{3}[/latex] 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, [latex]x>-\frac{4}{3}[/latex], 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 [latex]{\mathrm{log}}_{3}\left(\frac{7{x}^{2}+21x}{7x\left(x - 1\right)\left(x - 2\right)}\right)[/latex].
Using the Power Rule for Logarithms
We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}[/latex]? One method is as follows:
[latex]\begin{cases}{\mathrm{log}}_{b}\left({x}^{2}\right)\hfill & ={\mathrm{log}}_{b}\left(x\cdot x\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}x\hfill \\ \hfill & =2{\mathrm{log}}_{b}x\hfill \end{cases}[/latex]
Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,
[latex]\begin{cases}100={10}^{2}\hfill & \sqrt{3}={3}^{\frac{1}{2}}\hfill & \frac{1}{e}={e}^{-1}\hfill \end{cases}[/latex]
A General Note: The Power Rule for Logarithms
The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.
[latex]{\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M[/latex]
How To: Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.
- Express the argument as a power, if needed.
- Write the equivalent expression by multiplying the exponent times the logarithm of the base.
Example 3: Expanding a Logarithm with Powers
Expand [latex]{\mathrm{log}}_{2}{x}^{5}[/latex].
Solution
The argument is already written as a power, so we identify the exponent, 5, and the base, x, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.
[latex]{\mathrm{log}}_{2}\left({x}^{5}\right)=5{\mathrm{log}}_{2}x[/latex]
Example 4: Rewriting an Expression as a Power before Using the Power Rule
Expand [latex]{\mathrm{log}}_{3}\left(25\right)[/latex] using the power rule for logs.
Solution
Expressing the argument as a power, we get [latex]{\mathrm{log}}_{3}\left(25\right)={\mathrm{log}}_{3}\left({5}^{2}\right)[/latex].
Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.
[latex]{\mathrm{log}}_{3}\left({5}^{2}\right)=2{\mathrm{log}}_{3}\left(5\right)[/latex]
Example 5: Using the Power Rule in Reverse
Rewrite [latex]4\mathrm{ln}\left(x\right)[/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.
Solution
Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\mathrm{ln}\left(x\right)[/latex], we identify the factor, 4, as the exponent and the argument, x, as the base, and rewrite the product as a logarithm of a power:
[latex]4\mathrm{ln}\left(x\right)=\mathrm{ln}\left({x}^{4}\right)[/latex].
Try It 5
Rewrite [latex]2{\mathrm{log}}_{3}4[/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.
