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]

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 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 x and 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{array}{l}{\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]

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{array}{l}\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}{l}\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]

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]