Chapter 5, Problem 39RE

To determine

To find: The sum or difference form of the given logarithm.

Answer to Problem 39RE

The sum form of the given logarithm is ${\log }_{3}u+2{\log }_{3}v-{\log }_{3}w$ .

Explanation of Solution

Given:

The given logarithm is ${\log }_3\left(\frac{u{v}^2}{w}\right)$ .

The given condition is,

  $\begin{array}{l}u>0\\ v>0\\ w>0\end{array}$

Calculation:

Consider the given logarithm is,

  ${\log }_3\left(\frac{u{v}^2}{w}\right)$

Consider the property of logarithm is,

  ${\log }_a\left(\frac{b}{c}\right)={\log }_{a}b-{\log }_{a}c$

Then,

  ${\log }_3\left(\frac{u{v}^2}{w}\right)={\log }_{3}u{v}^2-{\log }_{3}w$

Consider the property of logarithm is,

  ${\log }_a\left(bc\right)={\log }_{a}b+{\log }_{a}c$

Then,

  ${\log }_3\left(\frac{u{v}^2}{w}\right)={\log }_{3}u+{\log }_3{v}^2-{\log }_{3}w$

Consider the property of logarithm is,

  ${\log }_a{b}^c=c{\log }_{a}b$

Then,

  ${\log }_3\left(\frac{u{v}^2}{w}\right)={\log }_{3}u+2{\log }_{3}v-{\log }_{3}w$