properties of logarithms assignment active

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• 8.1 Intro to Logs
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• 8.3 Properties of Logarithms
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8.3 Properties of Logaritms

Corrective Assignment

Module 12: Exponential and Logarithmic Equations and Models

Properties of logarithms, learning outcomes.

• 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 logarithms on the left side of the equation.

Using the Product Rule for Logarithms

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}{lllllllll}{\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]

Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]\mathrm{log}_{b}(wxyz)[/latex]. Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:

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].

We begin by writing an equal 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]

Expand [latex]{\mathrm{log}}_{b}\left(8k\right)[/latex].

[latex]{\mathrm{log}}_{b}8+{\mathrm{log}}_{b}k[/latex]

Using the Quotient Rule 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].

[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}{lllll}\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}{lll}\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]

A General Note: The Quotient Rule for Logarithms

The quotient rule for logarithms can be used to simplify a logarithm or 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]

How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms

• Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
• Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
• Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

Example: Using the Quotient Rule for Logarithms

Expand [latex]{\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)[/latex].

First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.

[latex]{\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)={\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right)[/latex]

Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule.

[latex]\begin{array}{l}{\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right) \\\text{}= \left[{\mathrm{log}}_{2}\left(15\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)\right]-\left[{\mathrm{log}}_{2}\left(3x+4\right)+{\mathrm{log}}_{2}\left(2-x\right)\right]\hfill \\ \text{}={\mathrm{log}}_{2}\left(15\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)-{\mathrm{log}}_{2}\left(3x+4\right)-{\mathrm{log}}_{2}\left(2-x\right)\hfill \end{array}[/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 x  = 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm that x  > 0, x  > 1, [latex]x>-\frac{4}{3}[/latex], and x  < 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

Expand [latex]{\mathrm{log}}_{3}\left(\frac{7{x}^{2}+21x}{7x\left(x - 1\right)\left(x - 2\right)}\right)[/latex].

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

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}{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}{lll}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

Rewrite [latex]{\mathrm{log}}_{2}{x}^{5}[/latex].

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]

Rewrite [latex]\mathrm{ln}{x}^{2}[/latex].

[latex]2\mathrm{ln}x[/latex]

Example: Rewriting an Expression as a Power before Using the Power Rule

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

Expressing the argument as a power, we get [latex]{\mathrm{log}}_{3}\left(25\right)={\mathrm{log}}_{3}\left({5}^{2}\right)[/latex].

Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

[latex]{\mathrm{log}}_{3}\left({5}^{2}\right)=2{\mathrm{log}}_{3}\left(5\right)[/latex]

Rewrite [latex]\mathrm{ln}\left(\frac{1}{{x}^{2}}\right)[/latex].

[latex]-2\mathrm{ln}\left(x\right)[/latex]

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7.4: Use the Properties of Logarithms

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Learning Objectives

By the end of this section, you will be able to:

• Use the properties of logarithms
• Use the Change of Base Formula

Before you get started, take this readiness quiz.

• Evaluate: a. \(a^{0}\) b. \(a^{1}\). If you missed this problem, review Example 5.14.
• Write with a rational exponent: \(\sqrt[3]{x^{2} y}\). If you missed this problem, review Example 8.27.
• Round to three decimal places: \(2.5646415\). If you missed this problem, review Example 1.34.

Use the Properties of Logarithms

Now that we have learned about exponential and logarithmic functions, we can introduce some of the properties of logarithms. These will be very helpful as we continue to solve both exponential and logarithmic equations.

The first two properties derive from the definition of logarithms. Since \(a^{0}=1\), we can convert this to logarithmic form and get \(\log _{a} 1=0\). Also, since \(a^{1}=a\), we get \(\log _{a} a=1\).

Definition \(\PageIndex{1}\)

Properties of logarithms.

• \(\log _{a} 1=0 \quad \log _{a} a=1\)

In the next example we could evaluate the logarithm by converting to exponential form, as we have done previously, but recognizing and then applying the properties saves time.

Example \(\PageIndex{1}\)

Evaluate using the properties of logarithms:

\(\log _{8} 1\)

\(\log _{6} 6\)

Use the property, \(\log _{a} 1=0\).

\(0 \quad \log _{8} 1=0\)

Use the property, \(\log _{a} a=1\).

\(1 \quad \log _{6} 6=1\)

Exercise \(\PageIndex{1}\)

• \(\log _{13} 1\)
• \(\log _{9} 9\)

Exercise \(\PageIndex{2}\)

• \(\log _{5} 1\)
• \(\log _{7} 7\)

The next two properties can also be verified by converting them from exponential form to logarithmic form, or the reverse.

The exponential equation \(a^{\log _{a} x}=x\) converts to the logarithmic equation \(\log _{a} x=\log _{a} x\), which is a true statement for positive values for \(x\) only.

The logarithmic equation \(\log _{a} a^{x}=x\) converts to the exponential equation \(a^{x}=a^{x}\), which is also a true statement.

These two properties are called inverse properties because, when we have the same base, raising to a power “undoes” the log and taking the log “undoes” raising to a power. These two properties show the composition of functions. Both ended up with the identity function which shows again that the exponential and logarithmic functions are inverse functions.

Definition \(\PageIndex{2}\)

Inverse Properties of Logarithms

For \(a>0, x>0\) and \(a \neq 1\),

\(a^{\log _{a} x}=x \quad \log _{a} a^{x}=x\)

In the next example, apply the inverse properties of logarithms.

Example \(\PageIndex{2}\)

\(4^{\log _{4} 9}\)

\(\log _{3} 3^{5}\)

Use the property, \(a^{\log _{a} x}=x\).

\(9 \quad 4^{\log _{4} 9}=9\)

\(5 \quad \log _{3} 3^{5}=5\)

Exercise \(\PageIndex{3}\)

• \(5^{\log _{5} 15}\)
• \(\log _{7} 7^{4}\)

Exercise \(\PageIndex{4}\)

• \(2^{\log _{2} 8}\)
• \(\log _{2} 2^{15}\)

There are three more properties of logarithms that will be useful in our work. We know exponential functions and logarithmic function are very interrelated. Our definition of logarithm shows us that a logarithm is the exponent of the equivalent exponential. The properties of exponents have related properties for exponents.

In the Product Property of Exponents, \(a^{m} \cdot a^{n}=a^{m+n}\), we see that to multiply the same base, we add the exponents. The Product Property of Logarithms , \(\log _{a} M \cdot N=\log _{a} M+\log _{a} N\) tells us to take the log of a product, we add the log of the factors.

Definition \(\PageIndex{3}\)

Product Property of Logarithms

If \(M>0, N>0, \mathrm{a}>0\) and \(\mathrm{a} \neq 1,\) then

\(\log _{a}(M \cdot N)=\log _{a} M+\log _{a} N\)

The logarithm of a product is the sum of the logarithms.

We use this property to write the log of a product as a sum of the logs of each factor.

Example \(\PageIndex{3}\)

Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible:

\(\log _{3} 7 x\)

\(\log _{4} 64 x y\)

Use the Product Property, \(\log _{a}(M \cdot N)=\log _{a} M+\log _{a} N\).

\(\log _{3} 7+\log _{3} x\) \(\log _{3} 7 x=\log _{3} 7+\log _{3} x\)

\(\log _{4} 64+\log _{4} x+\log _{4} y\)

Simplify be evaluating, \(\log _{4} 64\).

\(3+\log _{4} x+\log _{4} y\) \(\log _{4} 64 x y=3+\log _{4} x+\log _{4} y\)

Exercise \(\PageIndex{5}\)

• \(\log _{3} 3 x\)
• \(\log _{2} 8 x y\)
• \(1+\log _{3} x\)
• \(3+\log _{2} x+\log _{2} y\)

Exercise \(\PageIndex{6}\)

• \(\log _{9} 9 x\)
• \(\log _{3} 27 x y\)
• \(1+\log _{9} x\)
• \(3+\log _{3} x+\log _{3} y\)

Similarly, in the Quotient Property of Exponents, \(\frac{a^{m}}{a^{n}}=a^{m-n}\), we see that to divide the same base, we subtract the exponents. The Quotient Property of Logarithms , \(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\) tells us to take the log of a quotient, we subtract the log of the numerator and denominator.

Definition \(\PageIndex{4}\)

Quotient Property of Logarithms

\(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\)

The logarithm of a quotient is the difference of the logarithms.

Note that \(\log _{a} M=\log _{a} N \not=\log _{a}(M-N)\).

We use this property to write the log of a quotient as a difference of the logs of each factor.

Example \(\PageIndex{4}\)

Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.

\(\log _{5} \frac{5}{7}\)

\(\log \frac{x}{100}\)

Use the Quotient Property, \(\log _{a} \frac{M}{N}=\log _{a} M-\log _{a} N\).

\(\log _{5} 5-\log _{5} 7\)

\(1-\log _{5} 7\)

\(\log _{5} \frac{5}{7}=1-\log _{5} 7\)

\(\log x-\log 100\)

\(\log x-2\)

\(\log \frac{x}{100}=\log x-2\)

Exercise \(\PageIndex{7}\)

• \(\log _{4} \frac{3}{4}\)
• \(\log \frac{x}{1000}\)
• \(\log _{4} 3-1\)
• \(\log x-3\)

Exercise \(\PageIndex{8}\)

• \(\log _{2} \frac{5}{4}\)
• \(\log \frac{10}{y}\)
• \(\log _{2} 5-2\)
• \(1-\log y\)

The third property of logarithms is related to the Power Property of Exponents, \(\left(a^{m}\right)^{n}=a^{m \cdot n}\), we see that to raise a power to a power, we multiply the exponents. The Power Property of Logarithms , \(\log _{a} M^{p}=p \log _{a} M\) tells us to take the log of a number raised to a power, we multiply the power times the log of the number.

Definition \(\PageIndex{5}\)

Power Property of Logarithms

If \(M>0, \mathrm{a}>0, \mathrm{a} \neq 1\) and \(p\) is any real number then,

\(\log _{a} M^{p}=p \log _{a} M\)

The log of a number raised to a power as the product product of the power times the log of the number.

We use this property to write the log of a number raised to a power as the product of the power times the log of the number. We essentially take the exponent and throw it in front of the logarithm.

Example \(\PageIndex{5}\)

Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.

\(\log _{5} 4^{3}\)

\(\log x^{10}\)

Use the Power Property, \(\log _{a} M^{p}=p \log _{a} M\).

3 \(\log _{5} 4\)

\(\log _{5} 4^{3}=3 \log _{5} 4\)

\(10\log x\)

\(\log x^{10}=10 \log x\)

Exercise \(\PageIndex{9}\)

• \(\log _{7} 5^{4}\)
• \(\log x^{100}\)
• \(4\log _{7} 5\)
• 100\(\cdot \log x\)

Exercise \(\PageIndex{10}\)

• \(\log _{2} 3^{7}\)
• \(\log x^{20}\)
• \(7\log _{2} 3\)
• \(20\cdot \log x\)

We summarize the Properties of Logarithms here for easy reference. While the natural logarithms are a special case of these properties, it is often helpful to also show the natural logarithm version of each property.

Now that we have the properties we can use them to “expand” a logarithmic expression. This means to write the logarithm as a sum or difference and without any powers.

We generally apply the Product and Quotient Properties before we apply the Power Property.

Example \(\PageIndex{6}\)

Use the Properties of Logarithms to expand the logarithm \(\log _{4}\left(2 x^{3} y^{2}\right)\). Simplify, if possible.

Use the Product Property, \(\log _{a} M \cdot N=\log _{a} M+\log _{a} N\).

Use the Power Property, \(\log _{a} M^{p}=p \log _{a} M\), on the last two terms. Simplify.

Exercise \(\PageIndex{11}\)

Use the Properties of Logarithms to expand the logarithm \(\log _{2}\left(5 x^{4} y^{2}\right)\). Simplify, if possible.

\(\log _{2} 5+4 \log _{2} x+2 \log _{2} y\)

Exercise \(\PageIndex{12}\)

Use the Properties of Logarithms to expand the logarithm \(\log _{3}\left(7 x^{5} y^{3}\right)\). Simplify, if possible.

\(\log _{3} 7+5 \log _{3} x+3 \log _{3} y\)

When we have a radical in the logarithmic expression, it is helpful to first write its radicand as a rational exponent.

Example \(\PageIndex{7}\)

Use the Properties of Logarithms to expand the logarithm \(\log _{2} \sqrt[4]{\frac{x^{3}}{3 y^{2} z}}\). Simplify, if possible.

\(\log _{2} \sqrt[4]{\frac{x^{3}}{3 y^{2} z}}\)

Rewrite the radical with a rational exponent.

\(\log _{2}\left(\frac{x^{3}}{3 y^{2} z}\right)^{\frac{1}{4}}\)

\(\frac{1}{4} \log _{2}\left(\frac{x^{3}}{3 y^{2} z}\right)\)

Use the Quotient Property, \(\log _{a} M \cdot N=\log _{a} M-\log _{a} N\).

\(\frac{1}{4}\left(\log _{2}\left(x^{3}\right)-\log _{2}\left(3 y^{2} z\right)\right)\)

Use the Product Property, \(\log _{a} M \cdot N=\log _{a} M+\log _{a} N\), in the second term.

\(\frac{1}{4}\left(\log _{2}\left(x^{3}\right)-\left(\log _{2} 3+\log _{2} y^{2}+\log _{2} z\right)\right)\)

Use the Power Property, \(\log _{a} M^{p}=p \log _{a} M\), inside the parentheses.

\(\frac{1}{4}\left(3 \log _{2} x-\left(\log _{2} 3+2 \log _{2} y+\log _{2} z\right)\right)\)

Simplify by distributing.

\(\frac{1}{4}\left(3 \log _{2} x-\log _{2} 3-2 \log _{2} y-\log _{2} z\right)\)

\(\log _{2} \sqrt[4]{\frac{x^{3}}{3 y^{2} z}}=\frac{1}{4}\left(3 \log _{2} x-\log _{2} 3-2 \log _{2} y-\log _{2} z\right)\)

Exercise \(\PageIndex{13}\)

Use the Properties of Logarithms to expand the logarithm \(\log _{4} \sqrt[5]{\frac{x^{4}}{2 y^{3} z^{2}}}\). Simplify, if possible.

\(\frac{1}{5}\left(4 \log _{4} x-\frac{1}{2}-3 \log _{4} y-2 \log _{4} z\right)\)

Exercise \(\PageIndex{14}\)

Use the Properties of Logarithms to expand the logarithm \(\log _{3} \sqrt[3]{\frac{x^{2}}{5 y z}}\). Simplify, if possible.

\(\frac{1}{3}\left(2 \log _{3} x-\log _{3} 5-\log _{3} y-\log _{3} z\right)\)

The opposite of expanding a logarithm is to condense a sum or difference of logarithms that have the same base into a single logarithm. We again use the properties of logarithms to help us, but in reverse.

To condense logarithmic expressions with the same base into one logarithm, we start by using the Power Property to get the coefficients of the log terms to be one and then the Product and Quotient Properties as needed.

Example \(\PageIndex{8}\)

Use the Properties of Logarithms to condense the logarithm \(\log _{4} 3+\log _{4} x-\log _{4} y\). Simplify, if possible.

The log expressions all have the same base, \(4\).

The first two terms are added, so we use the Product Property, \(\log _{a} M+\log _{a} N=\log _{a} M : N\).

Since the logs are subtracted, we use the Quotient Property, \(\log _{a} M-\log _{a} N=\log _{a} \frac{M}{N}\).

Exercise \(\PageIndex{15}\)

Use the Properties of Logarithms to condense the logarithm \(\log _{2} 5+\log _{2} x-\log _{2} y\). Simplify, if possible.

\(\log _{2} \frac{5 x}{y}\)

Exercise \(\PageIndex{16}\)

Use the Properties of Logarithms to condense the logarithm \(\log _{3} 6-\log _{3} x-\log _{3} y\). Simplify, if possible.

\(\log _{3} \frac{6}{x y}\)

Example \(\PageIndex{9}\)

Use the Properties of Logarithms to condense the logarithm \(2 \log _{3} x+4 \log _{3}(x+1)\). Simplify, if possible.

The log expressions have the same base, \(3\).

\(2 \log _{3} x+4 \log _{3}(x+1)\)

Use the Power Property, \(\log _{a} M+\log _{a} N=\log _{a} M \cdot N\).

\(\log _{3} x^{2}+\log _{3}(x+1)^{4}\)

The terms are added, so we use the Product Property, \(\log _{a} M+\log _{a} N=\log _{a} M \cdot N\).

\(\log _{3} x^{2}(x+1)^{4}\) \(2 \log _{3} x+4 \log _{3}(x+1)=\log _{3} x^{2}(x+1)^{4}\)

Exercise \(\PageIndex{17}\)

Use the Properties of Logarithms to condense the logarithm \(3 \log _{2} x+2 \log _{2}(x-1)\). Simplify, if possible.

\(\log _{2} x^{3}(x-1)^{2}\)

Exercise \(\PageIndex{18}\)

Use the Properties of Logarithms to condense the logarithm \(2 \log x+2 \log (x+1)\). Simplify, if possible.

\(\log x^{2}(x+1)^{2}\)

Use the Change-of-Base Formula

To evaluate a logarithm with any other base, we can use the Change-of-Base Formula . We will show how this is derived.

\(\begin{array} {l c} {\text{Suppose we want to evaluate} \log_{a}M} & {\log_{a}M} \\ {\text{Let} \:y =\log_{a}M. }&{y=\log_{a}M} \\ {\text{Rewrite the epression in exponential form. }}&{a^{y}=M } \\ {\text{Take the }\:\log_{b} \text{of each side.}}&{\log_{b}a^{y}=\log_{b}M}\\ {\text{Use the Power Property.}}&{y\log_{b}a=\log_{b}M} \\ {\text{Solve for}\:y. }&{y=\frac{\log_{b}M}{\log_{b}a}} \\ {\text{Substiture}\:y=\log_{a}M.}&{\log_{a}M=\frac{\log_{b}M}{\log_{b}a}} \end{array}\)

The Change-of-Base Formula introduces a new base \(b\). This can be any base \(b\) we want where \(b>0,b≠1\). Because our calculators have keys for logarithms base \(10\) and base \(e\), we will rewrite the Change-of-Base Formula with the new base as \(10\) or \(e\).

Definition \(\PageIndex{6}\)

Change-of-Base Formula

For any logarithmic bases \(a, b\) and \(M>0\),

\(\begin{array}{lll}{\log _{a} M=\frac{\log _{b} M}{\log _{b} a}} & {\log _{a} M=\frac{\log M}{\log a}} & {\log _{a} M=\frac{\ln M}{\ln a}} \\ {\text { new base } b} & {\text { new base } 10} & {\text { new base } e}\end{array}\)

When we use a calculator to find the logarithm value, we usually round to three decimal places. This gives us an approximate value and so we use the approximately equal symbol \((≈)\).

Example \(\PageIndex{10}\)

Rounding to three decimal places, approximate \(\log _{4} 35\).

Exercise \(\PageIndex{19}\)

Rounding to three decimal places, approximate \(\log _{3} 42\).

Exercise \(\PageIndex{20}\)

Rounding to three decimal places, approximate \(\log _{5} 46\).

Access these online resources for additional instruction and practice with using the properties of logarithms.

• Using Properties of Logarithms to Expand Logs
• Using Properties of Logarithms to Condense Logs
• Change of Base

Key Concepts

\(\log _{a} M \cdot N=\log _{a} M+\log _{a} N\)

• Properties of Logarithms Summary If \(M>0,a>0,a≠1\) and \(p\) is any real number then,

\(\begin{array}{ll}{\log _{a} M=\frac{\log _{b} M}{\log _{b} a}} & {\log _{a} M=\frac{\log M}{\log a}} & {\log _{a} M=\frac{\ln M}{\ln a}} \\ {\text { new base } b} & {\text { new base } 10} & {\text { new base } e}\end{array}\)

Properties of Logarithms

When pre-calculus students learn about logarithmic functions, one of the most important lessons they come across is the properties of logarithms. This is because students can simplify and evaluate logarithms with the help of these properties.

Even though log lessons may be challenging, math teachers can help make them more engaging and accessible by using various teaching strategies. We share a few such strategies in this article. Read on and learn more!

Strategies for Teaching Properties of Logarithms

Review logarithms.

Start your lesson on the properties of logarithms by briefly reviewing what logarithms are. Remind students that a logarithm is an exponent. That is, log a x (“log base a of x”) is the exponent to which a must be raised to get x . You can present this in the following manner:

Where a > 0, a ≠ 1, and x > 0.

So logarithms are the opposite of exponentials, as they basically “undo” exponentials. You can also play this video in your class. The video introduces what logarithms represent, by using examples.

Then, check if there are any gaps in what students have learned so far on logarithms. For example, write a simple log on the whiteboard, such as log 2 16 = x, and ask students to transform it into an exponent. Can most students easily say that the equivalent exponent is 2 x = 16 ?

What about evaluating logarithms? Have students acquired proficiency in evaluating a given log? For instance, write log 3 81 = x on the whiteboard. Can students easily determine what this evaluates to, that is, log 3 81 = 4? Practice a bit more and address potential gaps.

For more advanced practice examples on evaluating logarithms, use this brief online activity by Khan Academy. If you require detailed guidelines on teaching logarithms, as well as fun activities to practice logarithms, feel free to check out this article.

Now that you’ve briefly reviewed them, you can proceed with explaining the properties of logarithms. For starters, highlight that we use the properties of logarithms for simplifying and evaluating logarithms.

Add that with the help of these properties, we can rewrite logarithmic expressions, that is, we can expand or condense them. Point out that you will look into three such properties in this class, including:

• multiplication property of logarithms
• division property of logarithms
• power rule of logarithms

Multiplication Property of Logarithms

Explain to students that the multiplication property of logarithms states that the logarithm of the product of two numbers is equal to the sum of individual logarithms of each number. Present this property on the whiteboard in the following way:

Division Property of Logarithms

Point out that according to the division property of logarithms, the logarithm of the quotient of two numbers is equal to the difference of the individual logarithm of each number. Present this property on the whiteboard in the following way:

Power Rule of Logarithms

Finally, explain that the power rule of logarithms states that the logarithm of a number raised to a certain power is equal to the product of power and logarithm of the number. Present this property on the whiteboard in the following way:

log 2 8 + log 2 32 = log 2 (8 × 32)

log 2 8 + log 2 32 = log 2 256

To check if this is correct, we can evaluate the logarithms, that is:

log 2 8 = 3, because 2 3 = 8

log 2 32 = 5, because 2 5 = 32

log 2 256 = 8, because 2 8 = 256

If we simply replace these values above in the statement log 2 8 + log 2 32 = log 2 256, we’ll get the following:

So there you have it! We see that the multiplication property is true.

Additional Resources:

If you have the technical possibilities, you can also complement your lesson with multimedia material, such as videos. For example, use this video by Khan Academy to introduce the multiplication, as well as the division property of logarithms.

Afterward, play this video by Khan Academy to illustrate the power rule of logarithms. By rewriting and simplifying log5(x3) as 3log5(x), the video demonstrates how the logarithmic power rule applies.

Activities to Practice Properties of Logarithms

Properties of logarithms game.

This is a simple online game that helps students sharpen their skills of simplifying and evaluating logarithms with the help of the properties of logarithms. To implement this game in your classroom, make sure that there is a sufficient number of devices for all students.

Students play the game individually, which makes the game suitable for parents who are homeschooling their kids, as well. Students are presented with different tasks, such as being asked to rewrite a log in a certain form by applying the properties of logarithms.

If they get stuck, students can also decide to play a video for help or use a hint. In the end, open space for discussion and reflection. Was any example particularly challenging? Why? Which properties did students use in their exercises?

This is a fun game that will help students improve their knowledge of logarithms and properties of logarithms. To play this game in your class, you’ll need to print out this free Log Race Worksheet , some dice, chips and scissors, markers, and some paper.

Print out as many copies as needed depending on the size of your class. Cut out the task cards from the worksheet and separate them into different piles, depending on what is required in them. For instance, the cards where students should expand a log go into a ‘log’ pile.

Then, draw a game board shaped like a road with designated spaces (squares), as shown on the worksheet, with a ‘start’ and ‘finish’ space. Each space (or square) on the road has a specific instruction, such as ‘expand’, ‘rewrite as exponent’, ‘evaluate’ etc.

Launch the Game!

Divide students into groups of 3, with one person as a ‘checker’. Player 1 rolls the dice and moves their chip on the game board the number of spaces that they got with the dice (ex: if they got a 3 by rolling the dice, they move their chip three spaces on the game board.

The space the student lands on with their chip indicates what kind of a card the student should take. For example, if the student landed on a space that says ‘condense’, they need to draw the top card from the ‘condense’ pile and condense the logarithmic expression written on the card.

Each student has a few minutes to solve the task on their card. If they solve it correctly, they get to roll the dice and move again. If they solve it incorrectly, they lose their turn in the next round. The designated checker in each group checks the answers and keeps track of the scores.

As the name of the game indicates, the goal is to be the first one to reach the ‘finish’ space. If no one manages to reach ‘finish’ by the end of the class, the winner is the player that was closest to ‘finish’.

Before You Leave…

If you enjoyed these tips and activities for teaching properties of logarithms, you may want to check out our lesson that’s dedicated to this topic. So if you need guidance to structure your class and teach pre-calculus, make sure to sign up for more free resources here !

Feel free to also check out our article with free resources on properties of logarithms!

You can also sign up for our membership on MathTeacherCoach or head over to our blog – you’ll find plenty of awesome resources for kids of all ages!

This article is based on:

Unit 3 – Exponential and Logarithmic Functions

• 3-1 Exponential Functions
• 3-2 Logarithmic Functions
• 3-3 Properties of Logarithms
• 3-4 Exponential and Logarithmic Equations
• 3-5 Modeling with Nonlinear Regression