Question

understanding properties of logarithms choose the letter of the expression listed in a through d 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}\\) = = = = 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:

To solve the problem of showing \(\log_b \frac{x}{y}\) using logarithmic properties, we analyze each step:

Step 1: Recall the Quotient Property of Logarithms

The quotient property states \(\log_b \frac{x}{y} = \log_b x - \log_b y\). However, we can also derive this using the power property (\(\log_b a^n = n\log_b a\)) and product property (\(\log_b (ab) = \log_b a + \log_b b\)).

First, rewrite \(\frac{x}{y}\) as \(x \cdot y^{-1}\) (since \(\frac{1}{y} = y^{-1}\)).

Step 2: Apply the Product Property

The product property of logarithms is \(\log_b (ab) = \log_b a + \log_b b\). For \(x \cdot y^{-1}\), this gives:
\[
\log_b (x \cdot y^{-1}) = \log_b x + \log_b y^{-1}
\]
This matches option A: \(\log_b x + \log_b y^{-1}\).

Step 3: Apply the Power Property

The power property states \(\log_b a^n = n\log_b a\). For \(\log_b y^{-1}\), this becomes:
\[
\log_b y^{-1} = -1 \cdot \log_b y = -\log_b y
\]
Substituting back, we get:
\[
\log_b x + \log_b y^{-1} = \log_b x - \log_b y
\]
This matches option B: \(\log_b x - \log_b y\).

Step 4: Verify the Final Form

The quotient property of logarithms confirms \(\log_b x - \log_b y = \log_b \frac{x}{y}\), which is the target expression.

Final Answer

To show \(\log_b \frac{x}{y}\), the steps use:

1. \(\boldsymbol{\log_b x + \log_b y^{-1}}\) (Option A)
2. \(\boldsymbol{\log_b x - \log_b y}\) (Option B)

(Note: The problem likely asks for the sequence of options to derive \(\log_b \frac{x}{y}\). The first step uses Option A, then simplifies to Option B, and so on.)

If selecting the first step (rewriting \(\frac{x}{y}\) as \(x \cdot y^{-1}\) and applying the product property), the answer is:

\(\boldsymbol{A. \log_b x + \log_b y^{-1}}\)

Answer:

To solve the problem of showing \(\log_b \frac{x}{y}\) using logarithmic properties, we analyze each step:

Step 1: Recall the Quotient Property of Logarithms

The quotient property states \(\log_b \frac{x}{y} = \log_b x - \log_b y\). However, we can also derive this using the power property (\(\log_b a^n = n\log_b a\)) and product property (\(\log_b (ab) = \log_b a + \log_b b\)).

First, rewrite \(\frac{x}{y}\) as \(x \cdot y^{-1}\) (since \(\frac{1}{y} = y^{-1}\)).

Step 2: Apply the Product Property

The product property of logarithms is \(\log_b (ab) = \log_b a + \log_b b\). For \(x \cdot y^{-1}\), this gives:
\[
\log_b (x \cdot y^{-1}) = \log_b x + \log_b y^{-1}
\]
This matches option A: \(\log_b x + \log_b y^{-1}\).

Step 3: Apply the Power Property

The power property states \(\log_b a^n = n\log_b a\). For \(\log_b y^{-1}\), this becomes:
\[
\log_b y^{-1} = -1 \cdot \log_b y = -\log_b y
\]
Substituting back, we get:
\[
\log_b x + \log_b y^{-1} = \log_b x - \log_b y
\]
This matches option B: \(\log_b x - \log_b y\).

Step 4: Verify the Final Form

The quotient property of logarithms confirms \(\log_b x - \log_b y = \log_b \frac{x}{y}\), which is the target expression.

Final Answer

To show \(\log_b \frac{x}{y}\), the steps use:

1. \(\boldsymbol{\log_b x + \log_b y^{-1}}\) (Option A)
2. \(\boldsymbol{\log_b x - \log_b y}\) (Option B)

(Note: The problem likely asks for the sequence of options to derive \(\log_b \frac{x}{y}\). The first step uses Option A, then simplifies to Option B, and so on.)

If selecting the first step (rewriting \(\frac{x}{y}\) as \(x \cdot y^{-1}\) and applying the product property), the answer is:

\(\boldsymbol{A. \log_b x + \log_b y^{-1}}\)