Precalculus Examples

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

Step 1

Apply the distributive property.

$\log_{8} (9) \log_{8} (x) + 7 \log_{8} (x)$

Step 2

Simplify $7 \log_{8} (x)$ by moving $7$ inside the logarithm.

$\log_{8} (9) \log_{8} (x) + \log_{8} (x^{7})$

Step 3

Simplify each term.

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Step 3.1

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

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Step 3.1.1

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 $0$ .

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

Step 3.1.2

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

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

Step 3.2

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

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Step 3.2.1

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 $0$ .

$\frac{\log (9)}{\log (8)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} + \log_{8} (x^{7})$

Step 3.2.2

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

$\frac{\log (9)}{\log (8)} \cdot \frac{\log (x)}{\log (8)} + \log_{8} (x^{7})$

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

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

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

Step 3.3.5

Add $1$ and $1$ .

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

Step 3.4

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

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Step 3.4.1

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 $0$ .

$\frac{\log (9) \log (x)}{\log^{2} (8)} + \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 3.4.2

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

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