Question

Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.

ⓐ ${\log }_{3}3x$ⓑ ${\log }_{2}8xy$

Short answer

ⓐ logarithm expression ${\log }_{3}3x$as the sum of logarithms isrole="math" localid="1644951471169" ${\log }_{3}3x=1+{\log }_{3}x$

ⓑ logarithm expression ${\log }_{2}8xy$as the sum of logarithms is ${\log }_{2}8xy=3+{\log }_{2}x+{\log }_{2}y$

Step-by-step solution

Step 1. given data 

Given logarithm expressions are

ⓐ${\log }_{3}3x$

ⓑ${\log }_{2}8xy$

Step 2. part (a) 

Use product Properties of Logarithms${\log }_a\left(m·n\right)={\log }_{a}m+{\log }_{a}n$

so$$\begin{gathered} {\log }_{3}3x={\log }_3(3·x) \\ {\log }_{3}3x={\log }_{3}3+{\log }_{3}x \\ {\log }_{3}3x=1+{\log }_{3}x \end{gathered}$$

Step 3. part (b) 

Use product Properties of Logarithms ${\log }_a\left(m·n\right)={\log }_{a}m+{\log }_{a}n$

sorole="math" localid="1644951807515" $$\begin{gathered} {\log }_{2}8xy={\log }_2(8·x·y) \\ {\log }_{2}8xy={\log }_{2}8+{\log }_{2}x+{\log }_{2}y \\ {\log }_{2}8xy={\log }_2{2}^3+{\log }_{2}x+{\log }_{2}y \\ {\log }_{2}8xy=3+{\log }_{2}x+{\log }_{2}y \end{gathered}$$