Question

In the following exercises, use the Properties of Logarithms to expand the logarithm. Simplify if possible.

${\log }_3\frac{\sqrt{3x+2y^2}}{5z^2}$

Short answer

The simplified expression is,

$\frac{1}{2}{\log }_3\left(3x+2y^2\right)-{\log }_{3}5+2{\log }_{3}z$

Step-by-step solution

Step 1. Given and explanation  

We are given an expression,

${\log }_3\frac{\sqrt{3x+2y^2}}{5z^2}$

Use the properties of logarithms to simplify the given expression.

Step 2. Simplifying the expression  

Using quotient property of logarithm , we get

$$\begin{gathered} {\log }_3\frac{\sqrt{3x+2y^2}}{5z^2}={\log }_3\sqrt{3x+2y^2}-{\log }_{3}5z^2 \\ {\log }_3\frac{{\displaystyle \sqrt{3x+2y^2}}}{{\displaystyle 5z^2}}={\log }_3{\left(3x+2y^2\right)}^{\frac{1}{2}}-{\log }_{3}5z^2 \end{gathered}$$

Step 3. Simplifying

Now, using product property of logarithm , we get

${\log }_3\frac{\sqrt{3x+2y^2}}{5z^2}={\log }_3{\left(3x+2y^2\right)}^{\frac{1}{2}}-\left({\log }_{3}5+{\log }_3{z}^2\right)$

Using power property of logarithm , we get

localid="1645015082469" ${\log }_3\frac{\sqrt{3x+2y^2}}{5z^2}=\frac{1}{2}{\log }_3\left(3x+2y^2\right)-{\log }_{3}5-2{\log }_{3}z$