Precalculus Examples

$\log_{8} (6) - \log_{8} (3) + \frac{1}{2} \cdot \log_{8} (4)$

Step 1

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

$\log_{8} (\frac{6}{3}) + \frac{1}{2} \cdot \log_{8} (4)$

Step 2

Simplify each term.

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Step 2.1

Logarithm base $8$ of $\frac{6}{3}$ is $\frac{1}{3}$ .

$\frac{1}{3} + \frac{1}{2} \cdot \log_{8} (4)$

Step 2.2

Logarithm base $8$ of $4$ is $\frac{2}{3}$ .

$\frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3}$

Step 2.3

Cancel the common factor of $2$ .

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Step 2.3.1

Cancel the common factor.

$\frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3}$

Step 2.3.2

Rewrite the expression.

$\frac{1}{3} + \frac{1}{3}$

Step 3

Combine fractions.

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Step 3.1

Combine the numerators over the common denominator.

$\frac{1 + 1}{3}$

Step 3.2

Add $1$ and $1$ .

$\frac{2}{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\frac{2}{3}$

Decimal Form:

$0.\overline{6}$