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Study Guides > College Algebra

Logarithmic Properties

Testing of the pH of hydrochloric acid. The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan)
In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:
  • Battery acid: 0.8
  • Stomach acid: 2.7
  • Orange juice: 3.3
  • Pure water: 7 (at 25° C)
  • Human blood: 7.35
  • Fresh coconut: 7.8
  • Sodium hydroxide (lye): 14
To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where a is the concentration of hydrogen ion in the solution

[latex]\begin{array}{l}\text{pH}=-\mathrm{log}\left(\left[{H}^{+}\right]\right)\hfill \\ \text{ }=\mathrm{log}\left(\frac{1}{\left[{H}^{+}\right]}\right)\hfill \end{array}[/latex]

The equivalence of [latex]-\mathrm{log}\left(\left[{H}^{+}\right]\right)[/latex] and [latex]\mathrm{log}\left(\frac{1}{\left[{H}^{+}\right]}\right)[/latex] is one of the logarithm properties we will examine in this section.

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.

  1. Express the argument as a power, if needed.
  2. 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]

Expand and Condense Logarithms

Taken together, the product rule, quotient rule, and power rule are often called "laws of logs." Sometimes we apply more than one rule in order to simplify an expression. For example:

[latex]\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{6x}{y}\right)\hfill & ={\mathrm{log}}_{b}\left(6x\right)-{\mathrm{log}}_{b}y\hfill \\ \hfill & ={\mathrm{log}}_{b}6+{\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y\hfill \end{array}[/latex]

We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power:

[latex]\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{A}{C}\right)\hfill & ={\mathrm{log}}_{b}\left(A{C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(A\right)+{\mathrm{log}}_{b}\left({C}^{-1}\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}A+\left(-1\right){\mathrm{log}}_{b}C\hfill \\ \hfill & ={\mathrm{log}}_{b}A-{\mathrm{log}}_{b}C\hfill \end{array}[/latex]

We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product. With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots—never with addition or subtraction inside the argument of the logarithm.

Example: Using a combination of the rules for logarithms to expand a logarithm

Rewrite [latex]\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)[/latex] as a sum or difference of logs.

Answer: First, because we have a quotient of two expressions, we can use the quotient rule: [latex-display]\mathrm{ln}\left(\frac{{x}^{4}y}{7}\right)=\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)[/latex-display] Then seeing the product in the first term, we use the product rule: [latex-display]\mathrm{ln}\left({x}^{4}y\right)-\mathrm{ln}\left(7\right)=\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)[/latex-display] Finally, we use the power rule on the first term: [latex-display]\mathrm{ln}\left({x}^{4}\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)=4\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)-\mathrm{ln}\left(7\right)[/latex-display]

Try It

Expand [latex]\mathrm{log}\left(\frac{{x}^{2}{y}^{3}}{{z}^{4}}\right)[/latex].

Answer: [latex-display]2\mathrm{log}x+3\mathrm{log}y - 4\mathrm{log}z[/latex-display]

Q & A

Can we expand [latex]\mathrm{ln}\left({x}^{2}+{y}^{2}\right)[/latex]?

No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm.
Now we will provide some examples that will require careful attention.

Example: Expanding Complex Logarithmic Expressions

Expand [latex]{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)[/latex].

Answer: We can expand by applying the Product and Quotient Rules. [latex-display]\begin{array}{l}{\mathrm{log}}_{6}\left(\frac{64{x}^{3}\left(4x+1\right)}{\left(2x - 1\right)}\right)\hfill & ={\mathrm{log}}_{6}64+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Quotient Rule}.\hfill \\ \hfill & ={\mathrm{log}}_{6}{2}^{6}+{\mathrm{log}}_{6}{x}^{3}+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & {\text{Simplify by writing 64 as 2}}^{6}.\hfill \\ \hfill & =6{\mathrm{log}}_{6}2+3{\mathrm{log}}_{6}x+{\mathrm{log}}_{6}\left(4x+1\right)-{\mathrm{log}}_{6}\left(2x - 1\right)\hfill & \text{Apply the Power Rule}.\hfill \end{array}[/latex-display]

Try It 8

Expand [latex]\mathrm{ln}\left(\frac{\sqrt{\left(x - 1\right){\left(2x+1\right)}^{2}}}{\left({x}^{2}-9\right)}\right)[/latex].

Answer: [latex]\frac{1}{2}\mathrm{ln}\left(x - 1\right)+\mathrm{ln}\left(2x+1\right)-\mathrm{ln}\left(x+3\right)-\mathrm{ln}\left(x - 3\right)[/latex]

Example: Condensing Complex Logarithmic Expressions

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

Answer: We apply the power rule first: [latex-display]{\mathrm{log}}_{2}\left({x}^{2}\right)+\frac{1}{2}{\mathrm{log}}_{2}\left(x - 1\right)-3{\mathrm{log}}_{2}\left({\left(x+3\right)}^{2}\right)={\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)[/latex-display] Next we apply the product rule to the sum: [latex-display]{\mathrm{log}}_{2}\left({x}^{2}\right)+{\mathrm{log}}_{2}\left(\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)[/latex-display] Finally, we apply the quotient rule to the difference: [latex-display]{\mathrm{log}}_{2}\left({x}^{2}\sqrt{x - 1}\right)-{\mathrm{log}}_{2}\left({\left(x+3\right)}^{6}\right)={\mathrm{log}}_{2}\frac{{x}^{2}\sqrt{x - 1}}{{\left(x+3\right)}^{6}}[/latex-display]

Example: Rewriting as a Single Logarithm

Rewrite [latex]2\mathrm{log}x - 4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)[/latex] as a single logarithm.

Answer: We apply the power rule first: [latex-display]2\mathrm{log}x - 4\mathrm{log}\left(x+5\right)+\frac{1}{x}\mathrm{log}\left(3x+5\right)=\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)[/latex-display] Next we apply the product rule to the sum: [latex-display]\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}\right)+\mathrm{log}\left({\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}{\left(3x+5\right)}^{{x}^{-1}}\right)[/latex-display] Finally, we apply the quotient rule to the difference: [latex-display]\mathrm{log}\left({x}^{2}\right)-\mathrm{log}\left({\left(x+5\right)}^{4}{\left(3x+5\right)}^{{x}^{-1}}\right)=\mathrm{log}\left(\frac{{x}^{2}}{{\left(x+5\right)}^{4}\left({\left(3x+5\right)}^{{x}^{-1}}\right)}\right)[/latex-display]

Try It

Rewrite [latex]\mathrm{log}\left(5\right)+0.5\mathrm{log}\left(x\right)-\mathrm{log}\left(7x - 1\right)+3\mathrm{log}\left(x - 1\right)[/latex] as a single logarithm.

Answer: [latex]\mathrm{log}\left(\frac{5{\left(x - 1\right)}^{3}\sqrt{x}}{\left(7x - 1\right)}\right)[/latex]

Condense [latex]4\left(3\mathrm{log}\left(x\right)+\mathrm{log}\left(x+5\right)-\mathrm{log}\left(2x+3\right)\right)[/latex].

Answer: [latex]\mathrm{log}\frac{{x}^{12}{\left(x+5\right)}^{4}}{{\left(2x+3\right)}^{4}}[/latex]; this answer could also be written [latex]\mathrm{log}{\left(\frac{{x}^{3}\left(x+5\right)}{\left(2x+3\right)}\right)}^{4}[/latex].

Change of Base

Use the change-of-base formula for logarithms

Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[/latex], we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs. To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms. Given any positive real numbers M, b, and n, where [latex]n\ne 1 [/latex] and [latex]b\ne 1[/latex], we show

[latex]{\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}[/latex]

Let [latex]y={\mathrm{log}}_{b}M[/latex]. By taking the log base [latex]n[/latex] of both sides of the equation, we arrive at an exponential form, namely [latex]{b}^{y}=M[/latex]. It follows that

[latex]\begin{array}{l}{\mathrm{log}}_{n}\left({b}^{y}\right)\hfill & ={\mathrm{log}}_{n}M\hfill & \text{Apply the one-to-one property}.\hfill \\ y{\mathrm{log}}_{n}b\hfill & ={\mathrm{log}}_{n}M \hfill & \text{Apply the power rule for logarithms}.\hfill \\ y\hfill & =\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\hfill & \text{Isolate }y.\hfill \\ {\mathrm{log}}_{b}M\hfill & =\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\hfill & \text{Substitute for }y.\hfill \end{array}[/latex]

For example, to evaluate [latex]{\mathrm{log}}_{5}36[/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.

[latex]\begin{array}{l}{\mathrm{log}}_{5}36\hfill & =\frac{\mathrm{log}\left(36\right)}{\mathrm{log}\left(5\right)}\hfill & \text{Apply the change of base formula using base 10}\text{.}\hfill \\ \hfill & \approx 2.2266\text{ }\hfill & \text{Use a calculator to evaluate to 4 decimal places}\text{.}\hfill \end{array}[/latex]

A General Note: The Change-of-Base Formula

The change-of-base formula can be used to evaluate a logarithm with any base. For any positive real numbers M, b, and n, where [latex]n\ne 1 [/latex] and [latex]b\ne 1[/latex],

[latex]{\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}[/latex].

It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.

[latex]{\mathrm{log}}_{b}M=\frac{\mathrm{ln}M}{\mathrm{ln}b}[/latex]

and

[latex]{\mathrm{log}}_{b}M=\frac{\mathrm{log}M}{\mathrm{log}b}[/latex]

How To: Given a logarithm with the form [latex]{\mathrm{log}}_{b}M[/latex], use the change-of-base formula to rewrite it as a quotient of logs with any positive base [latex]n[/latex], where [latex]n\ne 1[/latex].

  1. Determine the new base n, remembering that the common log, [latex]\mathrm{log}\left(x\right)[/latex], has base 10, and the natural log, [latex]\mathrm{ln}\left(x\right)[/latex], has base e.
  2. Rewrite the log as a quotient using the change-of-base formula
    • The numerator of the quotient will be a logarithm with base n and argument M.
    • The denominator of the quotient will be a logarithm with base n and argument b.

Example: Changing Logarithmic Expressions to Expressions Involving Only Natural Logs

Change [latex]{\mathrm{log}}_{5}3[/latex] to a quotient of natural logarithms.

Answer: Because we will be expressing [latex]{\mathrm{log}}_{5}3[/latex] as a quotient of natural logarithms, the new base, n = e. We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.

[latex]\begin{array}{l}{\mathrm{log}}_{b}M\hfill & =\frac{\mathrm{ln}M}{\mathrm{ln}b}\hfill \\ {\mathrm{log}}_{5}3\hfill & =\frac{\mathrm{ln}3}{\mathrm{ln}5}\hfill \end{array}[/latex]

Try It

Change [latex]{\mathrm{log}}_{0.5}8[/latex] to a quotient of natural logarithms.

Answer: [latex]\frac{\mathrm{ln}8}{\mathrm{ln}0.5}[/latex]

Key Equations

The Product Rule for Logarithms [latex]{\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)[/latex]
The Quotient Rule for Logarithms [latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N[/latex]
The Power Rule for Logarithms [latex]{\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M[/latex]
The Change-of-Base Formula [latex]{\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\text{ }n>0,n\ne 1,b\ne 1[/latex]

Key Concepts

  • We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms.
  • We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms.
  • We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base.
  • We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input.
  • The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm.
  • We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.
  • The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and e as the quotient of natural or common logs. That way a calculator can be used to evaluate.

Glossary

change-of-base formula a formula for converting a logarithm with any base to a quotient of logarithms with any other base. power rule for logarithms a rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base product rule for logarithms a rule of logarithms that states that the log of a product is equal to a sum of logarithms quotient rule for logarithms a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms

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