Question

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

ⓐ ${\log }_9\left(9x\right)$ⓑ ${\log }_3\left(27xy\right)$

Short answer

ⓐ logarithm expression ${\log }_9\left(9x\right)$as the sum of logarithms is ${\log }_9\left(9x\right)=1+{\log }_{9}x$

ⓑ logarithm expression ${\log }_3\left(27xy\right)$as the sum of logarithms is${\log }_3\left(27xy\right)=3+{\log }_{3}x+{\log }_{3}y$

Step-by-step solution

Step 1. given data 

Given logarithm expressions are

ⓐ${\log }_9\left(9x\right)$

ⓑ${\log }_3\left(27xy\right)$

Step 2. part (a) 

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

so$$\begin{gathered} {\log }_9\left(9x\right)={\log }_9\left(9·x\right) \\ {\log }_9\left(9x\right)={\log }_{9}9+{\log }_{9}x \\ {\log }_9\left(9x\right)=1+{\log }_{9}x \end{gathered}$$

Step 3. part (b) 

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

so

$$\begin{gathered} {\log }_3\left(27xy\right)={\log }_3\left(27·x·y\right) \\ {\log }_3\left(27xy\right)={\log }_{3}27+{\log }_{3}x+{\log }_{3}y \\ {\log }_3\left(27xy\right)={\log }_3{3}^3+{\log }_{3}x+{\log }_{3}y \\ {\log }_3\left(27xy\right)=3+{\log }_{3}x+{\log }_{3}y \end{gathered}$$