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8.8: Product and Quotient Properties of Logarithms

Difficulty Level: At Grade Created by: CK-12
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Practice Product and Quotient Properties of Logarithms
 
 
 
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Your friend Robbie works as a server at a pizza parlor. You and two of your friends go to the restaurant and order a pizza. You ask Robbie to bring you separate checks so you can split the cost of the pizza. Instead of bringing you three checks, Robbie brings you one with the total \begin{align*}\log_3 162 - \log_3 2\end{align*}log3162log32. "This is how much each of you owes," he says as he drops the bill on the table. How much do each of you owe?

Guidance

Just like exponents, logarithms have special properties, or shortcuts, that can be applied when simplifying expressions. In this lesson, we will address two of these properties.

Example A

Simplify \begin{align*}\log_b x + \log_b y\end{align*}logbx+logby.

Solution: First, notice that these logs have the same base. If they do not, then the properties do not apply.

\begin{align*}\log_b x=m\end{align*}logbx=m and \begin{align*}\log_b y=n\end{align*}logby=n, then \begin{align*}b^m=x\end{align*}bm=x and \begin{align*}b^n=y\end{align*}bn=y.

Now, multiply the latter two equations together.

\begin{align*}b^m \cdot b^n &= xy \\ b^{m+n} &= xy\end{align*}bmbnbm+n=xy=xy

Recall, that when two exponents with the same base are multiplied, we can add the exponents. Now, reapply the logarithm to this equation.

\begin{align*}b^{m+n}=xy \rightarrow \log_b xy=m+n\end{align*}bm+n=xylogbxy=m+n

Recall that \begin{align*}m=\log_b x\end{align*}m=logbx and \begin{align*}n=\log_b y\end{align*}n=logby, therefore \begin{align*}\log_b xy=\log_b x + \log_b y\end{align*}logbxy=logbx+logby.

This is the Product Property of Logarithms.

Example B

Expand \begin{align*}\log_{12} 4y\end{align*}log124y.

Solution: Applying the Product Property from Example A, we have:

\begin{align*}\log_{12} 4y = \log_{12} 4 + \log_{12} y\end{align*}log124y=log124+log12y

Example C

Simplify \begin{align*}\log_3 15 - \log_3 5\end{align*}log315log35.

Solution: As you might expect, the Quotient Property of Logarithms is \begin{align*}\log_b \frac{x}{y}=\log_b x - \log_b y\end{align*}logbxy=logbxlogby (proof in the Problem Set). Therefore, the answer is:

\begin{align*}\log_3 15 - \log_3 5 &= \log_3 \frac{15}{5} \\ &= \log_3 3 \\ &= 1\end{align*}log315log35=log3155=log33=1

Intro Problem Revisit

If you rewrite \begin{align*}\log_3 162 - \log_3 2\end{align*}log3162log32 as \begin{align*}\log_3 \frac {162}{2}\end{align*}log31622, you get \begin{align*}\log_3 81\end{align*}log381.

\begin{align*}3^4 = 81\end{align*}34=81 so you each owe $4.

Guided Practice

Simplify the following expressions.

1. \begin{align*}\log_7 8 + \log_7 x^2 + \log_7 3y\end{align*}log78+log7x2+log73y

2. \begin{align*}\log y - \log 20 + \log 8x\end{align*}logylog20+log8x

3. \begin{align*}\log_2 32 - \log_2 z\end{align*}log232log2z

4. \begin{align*}\log_8 \frac{16x}{y^2}\end{align*}log816xy2

Answers

1. Combine all the logs together using the Product Property.

\begin{align*}\log_7 8 + \log_7 x^2 + \log_7 3y &= \log_7 8x^2 3y \\ &= \log_7 24x^2 y\end{align*}log78+log7x2+log73y=log78x23y=log724x2y

2. Use both the Product and Quotient Property to condense.

\begin{align*}\log y - \log 20 + \log 8x &= \log \frac{y}{20} \cdot 8x \\ &= \log \frac{2xy}{5}\end{align*}logylog20+log8x=logy208x=log2xy5

3. Be careful; you do not have to use either rule here, just the definition of a logarithm.

\begin{align*}\log_2 32 - \log_2 z=5 - \log_2 z\end{align*}log232log2z=5log2z

4. When expanding a log, do the division first and then break the numerator apart further.

\begin{align*}\log_8 \frac{16x}{y^2} &= \log_8 16x - \log_8 y^2 \\ &= \log_8 16 + \log_8 x-\log_8 y^2 \\ &= \frac{4}{3} + \log_8 x - \log_8 y^2\end{align*}log816xy2=log816xlog8y2=log816+log8xlog8y2=43+log8xlog8y2

To determine \begin{align*}\log_8 16\end{align*}log816, use the definition and powers of 2: \begin{align*}8^n=16 \rightarrow 2^{3n}=2^4 \rightarrow 3n = 4 \rightarrow n=\frac{4}{3}\end{align*}8n=1623n=243n=4n=43.

Vocabulary

Product Property of Logarithms
As long as \begin{align*}b \ne 1\end{align*}b1, then \begin{align*}\log_b xy=\log_b x + \log_b y\end{align*}logbxy=logbx+logby
Quotient Property of Logarithms
As long as \begin{align*}b \ne 1\end{align*}b1, then \begin{align*}\log_b \frac{x}{y}=\log_b x - \log_b y\end{align*}logbxy=logbxlogby

Practice

Simplify the following logarithmic expressions.

  1. \begin{align*}\log_3 6 + \log_3 y - \log_3 4\end{align*}log36+log3ylog34
  2. \begin{align*}\log 12 - \log x + \log y^2\end{align*}log12logx+logy2
  3. \begin{align*}\log_6 x^2 - \log_6 x - \log_6 y\end{align*}log6x2log6xlog6y
  4. \begin{align*}\ln 8 + \ln 6 - \ln 12\end{align*}ln8+ln6ln12
  5. \begin{align*}\ln 7 - \ln 14 + \ln 10\end{align*}ln7ln14+ln10
  6. \begin{align*}\log_{11} 22 + \log_{11} 5 - \log_{11} 55\end{align*}log1122+log115log1155

Expand the following logarithmic functions.

  1. \begin{align*}\log_6 (5x)\end{align*}log6(5x)
  2. \begin{align*}\log_3 (abc)\end{align*}log3(abc)
  3. \begin{align*}\log \left(\frac{a^2}{b}\right)\end{align*}log(a2b)
  4. \begin{align*}\log_9 \left(\frac{xy}{5}\right)\end{align*}
  5. \begin{align*}\log \left(\frac{2x}{y}\right)\end{align*}
  6. \begin{align*}\log \left(\frac{8x^2}{15}\right)\end{align*}
  7. \begin{align*}\log_4 \left(\frac{5}{9y}\right)\end{align*}
  8. Write an algebraic proof of the Quotient Property. Start with the expression \begin{align*}\log_a x - \log_a y\end{align*} and the equations \begin{align*}\log_a x=m\end{align*} and \begin{align*}\log_a y=n\end{align*} in your proof. Refer to the proof of the product property in Example A as a guide for your proof.

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Vocabulary

Product Property of Logarithms

The product property of logarithms states that as long as b \ne 1, then \log_b xy=\log_b x + \log_b y

Quotient Property of Logarithms

The quotient property of logarithms states that as long as b \ne 1, then \log_b \frac{x}{y}=\log_b x - \log_b y.

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Mar 12, 2013
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MAT.ALY.334.1.L.1