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Study Guides > Intermediate Algebra

Read: Quotient and Power Rules for Logarithms

Learning Objectives

  • Define the quotient and power rules for logarithms
  • Use the quotient and power rules for logarithms to simplify logarithmic expressions

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.

 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]
We can show [latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{=}{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)[/latex]
Given positive real numbers M, N, and b, where [latex]b>0[/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{array}{c}{\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{array}[/latex]

Example

Expand the following expression using the quotient rule for logarithms. [latex-display]\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right)[/latex-display]

Answer:

Factoring and canceling we get,

[latex]\begin{array}{c}\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{array}[/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{array}{c}\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{array}[/latex]

In the previous example, it was helpful to first factor the numerator and denominator and divide common terms.  This gave us a simpler expression to use to write an equivalent expression. It is important to remember to subtract the logarithm of the denominator from the logarithm of the numerator. Always check to see if you need to expand further with the product rule.

Example

Expand [latex]{\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)[/latex].

Answer:

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 [latex]15[/latex] are [latex]3[/latex] and [latex]5[/latex].

[latex]\begin{array}{c}{\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{array}[/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 [latex]x = 2[/latex]. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that [latex]x > 0[/latex], [latex]x > 1[/latex], [latex]x>-\frac{4}{3}[/latex], and [latex]x < 2[/latex]. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

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{array}{c}{\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{array}[/latex]

Notice that we used the product rule for logarithms to simplify 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{array}{c}100={10}^{2}\hfill & \sqrt{3}={3}^{\frac{1}{2}}\hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}[/latex]

 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]

Example

Expand [latex]{\mathrm{log}}_{2}{x}^{5}[/latex].

Answer:

The argument is already written as a power, so we identify the exponent, [latex]5[/latex], 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]

The power rule for logarithms is possible because we can use the product rule and combine like terms. In the next example you will see that we can also rewrite an expression as a power in order to use the power rule.

Example

Expand [latex]{\mathrm{log}}_{3}\left(25\right)[/latex] using the power rule for logs.

Answer:

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, [latex]2[/latex], and the base, [latex]5[/latex], 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]

Now let's use the power rule in reverse.

Example

Rewrite [latex]4\mathrm{ln}\left(x\right)[/latex] using the power rule for logs to a single logarithm with a leading coefficient of [latex]1[/latex].

Answer:

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, [latex]4[/latex], 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].

 Summary

You can use the quotient rule of logarithms to write an equivalent difference of logarithms in the following way:

  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.

To use the power rule of logarithms to write an equivalent product of a factor and a logarithm consider the following:

  1. Express the argument as a power, if needed.
  2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.
 

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  • Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution. License terms: Download For Free at : http://cnx.org/contents/[email protected]..