$$ 1 \leftarrow \quad \begin{array}{l} ext { Use properties of logarithms to expand each logarithmic } \ ext { expression as much as possible. Evaluate logarithmic } \ ext { expressions without using a calculator if possible. } \ \log _{3}\left(\frac{\sqrt{y}}{9} ight) \ \log _{3}\left(\frac{\sqrt{y}}{9} ight)=\square\end{array} $$

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$$ 1 \leftarrow \quad \begin{array}{l}	ext { Use properties of logarithms to expand each logarithmic } \ 	ext { expression as much as possible. Evaluate logarithmic } \ 	ext { expressions without using a calculator if possible. } \ \log _{3}\left(\frac{\sqrt{y}}{9}ight) \ \log _{3}\left(\frac{\sqrt{y}}{9}ight)=\square\end{array} $$

UpStudy AI · Accepted Answer

We start by applying the quotient rule of logarithms:

$$
\log_{3}\left(\frac{\sqrt{y}}{9}\right) = \log_{3}\left(\sqrt{y}\right) - \log_{3}(9)
$$

Rewrite $$\sqrt{y}$$ as $$y^{\frac{1}{2}}$$ so that:

$$
\log_{3}\left(\sqrt{y}\right) = \frac{1}{2}\log_{3}(y)
$$

Recognize that $$9$$ can be written as $$3^2$$. Thus:

$$
\log_{3}(9) = \log_{3}\left(3^2\right) = 2
$$

Substituting these into the expression gives:

$$
\log_{3}\left(\frac{\sqrt{y}}{9}\right) = \frac{1}{2}\log_{3}(y) - 2
$$
$$ \log_{3}\left(\frac{\sqrt{y}}{9}\right) = \frac{1}{2}\log_{3}(y) - 2 $$

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