Quotient and Power Rules for Logarithms

Learning Outcome

  • Define and use the quotient and power rules for logarithms

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to simplify division of like bases raised to powers by subtracting the exponents: [latex]\frac{x^a}{x^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 with 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 and product rules for logarithms.

[latex]\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right)[/latex]

In the previous example, it was helpful to first factor the numerator and denominator and cancel common factors. 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 can 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].

 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\gt0[/latex], [latex]x\gt1[/latex], [latex]x\gt-\frac{4}{3}[/latex], and [latex]x\lt2[/latex]. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

In the following video, we show more examples of using the quotient rule for logarithms.

Using the Power Rule for Logarithms

We have 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.  The power rule for logarithms is possible because we can use the product rule and combine like terms.

 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].

Keep in mind that, even if the input to a logarithm is not written as a power, we may be able to change it to a power. You can see this demonstrated in the following three examples,

Example 1:  [latex]{\mathrm{log}}_{b}\left(100\right)={\mathrm{log}}_{b}\left({10}^{2}\right)[/latex]

Example 2:  [latex]{\mathrm{log}}_{b}\left(\sqrt{3}\right)={\mathrm{log}}_{b}\left({3}^{\frac{1}{2}}\right)[/latex]

Example 3:  [latex]{\mathrm{ln}}\left(\frac{1}{e}\right)={\mathrm{ln}}\left({e}^{-1}\right)[/latex]

In the next example, we will rewrite an expression as a power, making it possible for us to use the power rule.

Example

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

Now let us use the power rule in reverse.

Example

Using the power rule for logarithms, rewrite [latex]4\mathrm{ln}\left(x\right)[/latex] as a single logarithm with a leading coefficient of [latex]1[/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.