Precalculus Examples

$\frac{1}{8} \cdot \log_{3} (x) - \log_{3} (y) + 9 \log_{3} (x + 5)$

Step 1

Simplify each term.

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Step 1.1

Simplify $\frac{1}{8} \log_{3} (x)$ by moving $\frac{1}{8}$ inside the logarithm.

$\log_{3} (x^{\frac{1}{8}}) - \log_{3} (y) + 9 \log_{3} (x + 5)$

Step 1.2

Simplify $9 \log_{3} (x + 5)$ by moving $9$ inside the logarithm.

$\log_{3} (x^{\frac{1}{8}}) - \log_{3} (y) + \log_{3} ({(x + 5)}^{9})$

Step 2

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

$\log_{3} (\frac{x^{\frac{1}{8}}}{y}) + \log_{3} ({(x + 5)}^{9})$

Step 3

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

$\log_{3} (\frac{x^{\frac{1}{8}}}{y} {(x + 5)}^{9})$

Step 4

Combine $\frac{x^{\frac{1}{8}}}{y}$ and ${(x + 5)}^{9}$ .

$\log_{3} (\frac{x^{\frac{1}{8}} {(x + 5)}^{9}}{y})$

Step 5

Rewrite $\log_{3} (\frac{x^{\frac{1}{8}} {(x + 5)}^{9}}{y})$ using the change of base formula.

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Step 5.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 5.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\log (\frac{x^{\frac{1}{8}} {(x + 5)}^{9}}{y})}{\log (3)}$