Logarithm Rules
Learning Objectives
- Rewrite a logarithmic expression using the power rule, product rule, or quotient rule
- Expand logarithmic expressions using a combination of logarithm rules
- Condense logarithmic expressions using logarithm rules
Recall that the logarithmic and exponential functions "undo" each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.
[latex]\begin{array}{l}{\mathrm{log}}_{b}1=0\\{\mathrm{log}}_{b}b=1\end{array}[/latex]
For example, [latex]{\mathrm{log}}_{5}1=0[/latex] since [latex]{5}^{0}=1[/latex]. And [latex]{\mathrm{log}}_{5}5=1[/latex] since [latex]{5}^{1}=5[/latex]. Next, we have the inverse property.[latex]\begin{array}{l}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}[/latex]
For example, to evaluate [latex]\mathrm{log}\left(100\right)[/latex], we can rewrite the logarithm as [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)[/latex], and then apply the inverse property [latex]{\mathrm{log}}_{b}\left({b}^{x}\right)=x[/latex] to get [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)=2[/latex]. To evaluate [latex]{e}^{\mathrm{ln}\left(7\right)}[/latex], we can rewrite the logarithm as [latex]{e}^{{\mathrm{log}}_{e}7}[/latex], and then apply the inverse property [latex]{b}^{{\mathrm{log}}_{b}x}=x[/latex] to get [latex]{e}^{{\mathrm{log}}_{e}7}=7[/latex]. Finally, we have the one-to-one property.[latex]{\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N[/latex]
We can use the one-to-one property to solve the equation [latex]{\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right)[/latex] for x. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x:[latex]\begin{array}{l}3x=2x+5\hfill & \text{Set the arguments equal}\text{.}\hfill \\ x=5\hfill & \text{Subtract 2}x\text{.}\hfill \end{array}[/latex]
But what about 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. Recall that we use the product rule of exponents to combine the product of exponents by adding: [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. Given any real number x and positive real numbers M, N, and b, where [latex]b\ne 1[/latex], we will show[latex]{\mathrm{log}}_{b}\left(MN\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}{l}{\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]
A General Note: 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]
Example: Using the Product Rule for Logarithms
Expand [latex]{\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)[/latex].Answer: We begin by writing the equivalent equation by summing the logarithms of each factor.
[latex]{\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)={\mathrm{log}}_{3}\left(30x\right)+{\mathrm{log}}_{3}\left(3x+4\right)={\mathrm{log}}_{3}\left(30\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)[/latex]
The final expansion looks like this, note how the factor [latex]30x[/latex] can be expanded into the sum of two logarithms:
[latex]{\mathrm{log}}_{3}\left(30\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)[/latex]
Try It
Expand [latex]{\mathrm{log}}_{b}\left(8k\right)[/latex].Answer: [latex]{\mathrm{log}}_{b}8+{\mathrm{log}}_{b}k[/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{array}{l}{\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 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{array}{l}100={10}^{2}\hfill & \sqrt{3}={3}^{\frac{1}{2}}\hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}[/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: Expanding a Logarithm with Powers
Expand [latex]{\mathrm{log}}_{2}{x}^{5}[/latex].Answer: 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]
Try It
Expand [latex]\mathrm{ln}{x}^{2}[/latex].Answer: [latex]2\mathrm{ln}x[/latex]
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- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
