Question

choose the letter of the expression listed on the right that completes each step to show how to use the power and product properties of logarithms to prove that the quotient property is true for $\log_{b}\frac{x}{y}$. $\log_{b}\frac{x}{y}$ $= \square$ $= \square$ $= \square$ $= \square$ a $\log_{b}x + \log_{b}y^{-1}$ b $\log_{b}x - \log_{b}y$ c $\log_{b}xy^{-1}$ d $\log_{b}x - 1\log_{b}y$

Explanation:

Step1: Rewrite the quotient as a product

Recall that \(\frac{x}{y}=xy^{-1}\). So, \(\log_{b}\frac{x}{y}=\log_{b}xy^{-1}\) (This is option C).

Step2: Apply the product property of logarithms

The product property of logarithms states that \(\log_{b}(MN)=\log_{b}M+\log_{b}N\). So, \(\log_{b}xy^{-1}=\log_{b}x+\log_{b}y^{-1}\) (This is option A).

Step3: Apply the power property of logarithms

The power property of logarithms states that \(\log_{b}M^{n}=n\log_{b}M\). So, \(\log_{b}y^{-1}=- 1\times\log_{b}y\), and \(\log_{b}x+\log_{b}y^{-1}=\log_{b}x - 1\log_{b}y\) (This is option D).

Step4: Simplify the expression

\(\log_{b}x-1\log_{b}y=\log_{b}x - \log_{b}y\) (This is option B).

Answer:

First step: C. \(\log_{b}xy^{-1}\)
Second step: A. \(\log_{b}x+\log_{b}y^{-1}\)
Third step: D. \(\log_{b}x - 1\log_{b}y\)
Fourth step: B. \(\log_{b}x-\log_{b}y\)