Precalculus Examples

$\frac{1}{3} \cdot \log_{a} (x) + \frac{1}{2} \cdot \log_{a} (2 y) - \frac{1}{6} \cdot \log_{a} (4 z)$

Step 1

Simplify each term.

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Step 1.1

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

$\log_{a} (x^{\frac{1}{3}}) + \frac{1}{2} \cdot \log_{a} (2 y) - \frac{1}{6} \cdot \log_{a} (4 z)$

Step 1.2

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

$\log_{a} (x^{\frac{1}{3}}) + \log_{a} ({(2 y)}^{\frac{1}{2}}) - \frac{1}{6} \cdot \log_{a} (4 z)$

Step 1.3

Apply the product rule to $2 y$ .

$\log_{a} (x^{\frac{1}{3}}) + \log_{a} (2^{\frac{1}{2}} y^{\frac{1}{2}}) - \frac{1}{6} \cdot \log_{a} (4 z)$

Step 1.4

Multiply $- \frac{1}{6} \log_{a} (4 z)$ .

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Step 1.4.1

Reorder $\log_{a} (4 z)$ and $\frac{1}{6}$ .

$\log_{a} (x^{\frac{1}{3}}) + \log_{a} (2^{\frac{1}{2}} y^{\frac{1}{2}}) - (\frac{1}{6} \log_{a} (4 z))$

Step 1.4.2

Simplify $\frac{1}{6} \log_{a} (4 z)$ by moving $\frac{1}{6}$ inside the logarithm.

$\log_{a} (x^{\frac{1}{3}}) + \log_{a} (2^{\frac{1}{2}} y^{\frac{1}{2}}) - \log_{a} ({(4 z)}^{\frac{1}{6}})$

Step 1.5

Apply the product rule to $4 z$ .

$\log_{a} (x^{\frac{1}{3}}) + \log_{a} (2^{\frac{1}{2}} y^{\frac{1}{2}}) - \log_{a} (4^{\frac{1}{6}} z^{\frac{1}{6}})$

Step 2

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

$\log_{a} (x^{\frac{1}{3}} (2^{\frac{1}{2}} y^{\frac{1}{2}})) - \log_{a} (4^{\frac{1}{6}} z^{\frac{1}{6}})$

Step 3

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

$\log_{a} (\frac{x^{\frac{1}{3}} \cdot 2^{\frac{1}{2}} y^{\frac{1}{2}}}{4^{\frac{1}{6}} z^{\frac{1}{6}}})$