Question

sam is proving the product property of logarithms.

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$$\begin{tabular}{|l|l|} \\hline step & justification \\\\ \\hline \\(\\log_b(mn)\\) & given \\\\ \\hline \\(=\\log_b(b^x \\cdot b^y)\\) & substitution \\\\ \\hline & \\\\ \\hline \\end{tabular}$$

which expression and justification completes the third step of her proof?

\\(\bigcirc\\) \\(\log_b(b^{xy})\\); power rule of exponents
\\(\bigcirc\\) \\(\log_b(b^{x-y})\\); subtraction property of exponents
\\(\bigcirc\\) \\(\log_b(b^{x+y})\\); multiplication rule of exponents
\\(\bigcirc\\) \\(\log_b(b^{\frac{x}{y}})\\); division property of exponents

Explanation:

⚡ Using what you learned: properties of logarithms

Step 1: Analyze the current step

The second step of the proof is:

$$ \log_b(b^x \cdot b^y) $$

Step 2: Apply exponent rules

To simplify the expression inside the logarithm, use the product rule of exponents (also referred to here as the multiplication rule of exponents):

$$ b^x \cdot b^y = b^{x+y} $$

Substituting this back into the logarithm gives:

$$ \log_b(b^{x+y}) $$

Step 3: Match with the options

The expression is \( \log_b(b^{x+y}) \) and the justification is the multiplication rule of exponents.

Answer:

\( \log_b(b^{x+y}) \); multiplication rule of exponents