Product Rule for Logarithms

Learning Outcomes

  • Define properties of logarithms, and use them to solve equations
  • Define and use the product rule for logarithms

Recall that we can express the relationship between logarithmic form and its corresponding exponential form as follows:

[latex]{\mathrm{log}}_{b}\left(x\right)=y\Leftrightarrow {b}^{y}=x,\text{}b>0,b\ne 1[/latex]

Note that the base b is always positive and that the logarithmic and exponential functions “undo” each other. This means that logarithms have properties similar to exponents. Some important properties of logarithms are given in this section.

First, we will introduce some basic properties of logarithms followed by examples with integer arguments to help you get familiar with the relationship between exponents and logarithms.

Zero and Identity Exponent Rule for Logarithms and Exponentials

[latex]{\mathrm{log}}_{b}1=0[/latex], b>[latex]0[/latex]
[latex]{\mathrm{log}}_{b}b=1[/latex], b>[latex]0[/latex]

Example

Use the the fact that exponentials and logarithms are inverses to prove the zero and identity exponent rule for the following:

1. [latex]{\mathrm{log}}_{5}1=0[/latex]

2. [latex]{\mathrm{log}}_{5}5=1[/latex]

Exponential and logarithmic functions are inverses of each other, and we can take advantage of this to evaluate and solve expressions and equations involving logarithms and exponentials. The inverse property of logarithms and exponentials gives us an explicit way to rewrite an exponential as a logarithm or a logarithm as an exponential.

Inverse Property of Logarithms and Exponentials

[latex]\begin{array}{c}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x, x>0, b>0, b\ne1\hfill \end{array}[/latex]

Example

Evaluate:

1.[latex]\mathrm{log}\left(100\right)[/latex]

2.[latex]{e}^{\mathrm{ln}\left(7\right)}[/latex]

Another property that can help us simplify logarithms is the one-to-one property. Essentially, this property states that if two logarithms have the same base, then their arguments – the stuff inside – are also equal to each other.

The One-To-One Property of Logarithms

[latex]{\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N[/latex]

Example

Solve the equation [latex]{\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right)[/latex] for [latex]x[/latex].

What if we had started with the equation [latex]{\mathrm{log}}_{3}\left(3x\right)+{\mathrm{log}}_{3}\left(2x+5\right)=2[/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.

To recap, the properties of logarithms and exponentials that can help us understand, simplify, and solve these types of functions more easily include:

  • Zero and Identity Exponent Rule: [latex]{\mathrm{log}}_{b}1=0[/latex], [latex]b\gt0[/latex] and [latex]{\mathrm{log}}_{b}b=1[/latex], [latex]b\gt0[/latex]
  • Inverse Property: [latex]\begin{array}{c}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}[/latex]
  • One-To-One Property: [latex]{\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N[/latex]

The Product Rule for Logarithms

Recall that we use the product rule of exponents to combine the product of like bases raised to exponents by adding the exponents: [latex]{x}^{a}{x}^{b}={x}^{a+b}[/latex]. We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.

The Product Rule for Logarithms

The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.

[latex]{\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\text{ for }b>0[/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(MN\right)\hfill & ={\mathrm{log}}_{b}\left({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 product 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]

Repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. Consider the following example:

Example

Using the product rule for logarithms, rewrite the logarithm of a product as the sum of logarithms of its factors.

[latex]{\mathrm{log}}_{b}\left(wxyz\right)[/latex]

In our next example, we will first factor the argument of a logarithm before expanding it with the product rule.

Example

Expand [latex]{\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)[/latex].

Analysis of the Solution

It is tempting to use the distributive property when you see an expression like [latex]\left(30x\left(3x+4\right)\right)[/latex], but in this case, it is better to leave the argument of this logarithm as a product since you can then use the product rule for logarithms to simplify the expression.

The following video provides more examples of using the product rule to expand logarithms.

Summary

Logarithms have properties that can help us simplify and solve expressions and equations that contain logarithms. Exponentials and logarithms are inverses of each other; therefore, we can define the product rule for logarithms. We can use this to simplify or solve expressions with logarithms. Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms as follows:

  1. Factor the argument completely, expressing each whole number factor as a product of primes.
  2. Write the equivalent expression by summing the logarithms of each factor.