Sure, let's break down the process step-by-step by using the power and product properties of logarithms to prove the quotient property for [tex]\(\log_b \frac{x}{y}\)[/tex].
### Step 1:
Start with the given equation:
[tex]\[
\log_b \frac{x}{y}
\][/tex]
### Step 2:
Use the power property of logarithms to express the division inside the logarithm as a multiplication with the exponent -1.
[tex]\[
\log_b \frac{x}{y} = \log_b (x \cdot y^{-1})
\][/tex]
Choose the correct option:
C [tex]\(\log_b (x \cdot y^{-1})\)[/tex]
### Step 3:
Express the logarithm of a product as the sum of the logarithms:
[tex]\[
\log_b (x \cdot y^{-1}) = \log_b x + \log_b y^{-1}
\][/tex]
Choose the correct option:
A [tex]\(\log_b x + \log_b y^{-1}\)[/tex]
### Step 4:
Use the power property of logarithms again to bring the exponent -1 in front of the logarithm:
[tex]\[
\log_b x + \log_b y^{-1} = \log_b x + (-\log_b y)
\][/tex]
Which simplifies to:
[tex]\[
\log_b x - \log_b y
\][/tex]
Choose the correct option:
B [tex]\(\log_b x - \log_b y\)[/tex]
### Step 5:
Simplify the expression to its final form:
[tex]\[
\log_b x - \log_b y
\][/tex]
Choose the correct option:
B [tex]\(\log_b x - \log_b y\)[/tex]
Putting it all together, we have:
[tex]\[
\begin{aligned}
\log_b \frac{x}{y} & = \log_b (x \cdot y^{-1}) \quad \text{(C)} \checkmark \\
& = \log_b x + \log_b y^{-1} \quad \text{(A)} \checkmark \\
& = \log_b x + (-\log_b y) \quad \text{(B)} \checkmark \\
& = \log_b x - \log_b y \quad \text{(B)} \checkmark
\end{aligned}
\][/tex]
Hence, the final proof that the quotient property is true for [tex]\(\log_b \frac{x}{y}\)[/tex] is complete.
