Precalculus Examples

$\log_{8} (7) \cdot \log_{8} (9)$

Step 1

Rewrite $\log_{8} (7)$ using the change of base formula.

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Step 1.1

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 $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{8} (9)$

Step 1.2

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

$\frac{\log (7)}{\log (8)} \log_{8} (9)$

Step 2

Rewrite $\log_{8} (9)$ using the change of base formula.

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Step 2.1

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 $0$ .

$\frac{\log (7)}{\log (8)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 2.2

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

$\frac{\log (7)}{\log (8)} \cdot \frac{\log (9)}{\log (8)}$

Step 3

Multiply $\frac{\log (7)}{\log (8)} \cdot \frac{\log (9)}{\log (8)}$ .

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Step 3.1

Multiply $\frac{\log (7)}{\log (8)}$ by $\frac{\log (9)}{\log (8)}$ .

$\frac{\log (7) \log (9)}{\log (8) \log (8)}$

Step 3.2

Raise $\log (8)$ to the power of $1$ .

$\frac{\log (7) \log (9)}{\log^{1} (8) \log (8)}$

Step 3.3

Raise $\log (8)$ to the power of $1$ .

$\frac{\log (7) \log (9)}{\log^{1} (8) \log^{1} (8)}$

Step 3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\log (7) \log (9)}{{\log (8)}^{1 + 1}}$

Step 3.5

Add $1$ and $1$ .

$\frac{\log (7) \log (9)}{\log^{2} (8)}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (7) \log (9)}{\log^{2} (8)}$

Decimal Form:

$0.98878939 \dots$