Precalculus Examples

$\frac{1}{5} \cdot \log_{3} (32) - 2 \log_{3} (x) + \frac{1}{2} \cdot \log_{3} (y)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{1}{5} \log_{3} (32)$ by moving $\frac{1}{5}$ inside the logarithm.

$\log_{3} (32^{\frac{1}{5}}) - 2 \log_{3} (x) + \frac{1}{2} \cdot \log_{3} (y)$

Step 1.2

Rewrite $32$ as $2^{5}$ .

$\log_{3} ({(2^{5})}^{\frac{1}{5}}) - 2 \log_{3} (x) + \frac{1}{2} \cdot \log_{3} (y)$

Step 1.3

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\log_{3} (2^{5 (\frac{1}{5})}) - 2 \log_{3} (x) + \frac{1}{2} \cdot \log_{3} (y)$

Step 1.4

Cancel the common factor of $5$ .

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Step 1.4.1

Cancel the common factor.

$\log_{3} (2^{5 (\frac{1}{5})}) - 2 \log_{3} (x) + \frac{1}{2} \cdot \log_{3} (y)$

Step 1.4.2

Rewrite the expression.

$\log_{3} (2^{1}) - 2 \log_{3} (x) + \frac{1}{2} \cdot \log_{3} (y)$

Step 1.5

Evaluate the exponent.

$\log_{3} (2) - 2 \log_{3} (x) + \frac{1}{2} \cdot \log_{3} (y)$

Step 1.6

Simplify $- 2 \log_{3} (x)$ by moving $2$ inside the logarithm.

$\log_{3} (2) - \log_{3} (x^{2}) + \frac{1}{2} \cdot \log_{3} (y)$

Step 1.7

Simplify $\frac{1}{2} \log_{3} (y)$ by moving $\frac{1}{2}$ inside the logarithm.

$\log_{3} (2) - \log_{3} (x^{2}) + \log_{3} (y^{\frac{1}{2}})$

Step 2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{3} (\frac{2}{x^{2}}) + \log_{3} (y^{\frac{1}{2}})$

Step 3

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{3} (\frac{2}{x^{2}} y^{\frac{1}{2}})$

Step 4

Combine $\frac{2}{x^{2}}$ and $y^{\frac{1}{2}}$ .

$\log_{3} (\frac{2 y^{\frac{1}{2}}}{x^{2}})$