Question

the proof for the product property of logarithms requires simplifying the expression $\log_{b}(b^{x + y})$ to $x + y$. which property is used to justify this step? $b^{x} cdot b^{y} = b^{x + y}$ substitution $\log_{b}(b^{c}) = c$ commutative property

Explanation:

To simplify \( \log_b(b^{x + y}) \) to \( x + y \), we use the property of logarithms that states \( \log_b(b^c)=c \). Let's analyze each option:

• The first option \( b^x \cdot b^y = b^{x + y} \) is the product property of exponents, not related to simplifying the logarithm here.
• Substitution is a general method but not a specific property for this logarithm simplification.
• The property \( \log_b(b^c)=c \) directly applies here. If we let \( c=x + y \), then \( \log_b(b^{x + y})=x + y \) by this property.
• The commutative property deals with the order of addition or multiplication (e.g., \( a + b=b + a \) or \( ab = ba \)) and is not relevant to this logarithm simplification.

Answer:

\( \boldsymbol{\log_b(b^c) = c} \) (the option with this property)