Precalculus Examples

$8^{\log_{8} (x^{2} - 3) (x + 2)}$

Step 1

Apply the distributive property.

$8^{\log_{8} (x^{2} - 3) x + \log_{8} (x^{2} - 3) \cdot 2}$

Step 2

Multiply $\log_{8} (x^{2} - 3) \cdot 2$ .

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Step 2.1

Reorder $\log_{8} (x^{2} - 3)$ and $2$ .

$8^{\log_{8} (x^{2} - 3) x + 2 \cdot \log_{8} (x^{2} - 3)}$

Step 2.2

Simplify $2 \cdot \log_{8} (x^{2} - 3)$ by moving $2$ inside the logarithm.

$8^{\log_{8} (x^{2} - 3) x + \log_{8} ({(x^{2} - 3)}^{2})}$

Step 3

Reorder factors in $8^{\log_{8} (x^{2} - 3) x + \log_{8} ({(x^{2} - 3)}^{2})}$ .

$8^{x \log_{8} (x^{2} - 3) + \log_{8} ({(x^{2} - 3)}^{2})}$

Step 4

Simplify each term.

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Step 4.1

Rewrite $\log_{8} (x^{2} - 3)$ using the change of base formula.

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Step 4.1.1

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 $0$ .

$8^{x \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} + \log_{8} ({(x^{2} - 3)}^{2})}$

Step 4.1.2

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

$8^{x \frac{\log (x^{2} - 3)}{\log (8)} + \log_{8} ({(x^{2} - 3)}^{2})}$

Step 4.2

Combine $x$ and $\frac{\log (x^{2} - 3)}{\log (8)}$ .

$8^{\frac{x \log (x^{2} - 3)}{\log (8)} + \log_{8} ({(x^{2} - 3)}^{2})}$

Step 4.3

Rewrite $\log_{8} ({(x^{2} - 3)}^{2})$ using the change of base formula.

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Step 4.3.1

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 $0$ .

$8^{\frac{x \log (x^{2} - 3)}{\log (8)} + \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}}$

Step 4.3.2

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

$8^{\frac{x \log (x^{2} - 3)}{\log (8)} + \frac{\log ({(x^{2} - 3)}^{2})}{\log (8)}}$

Step 5

Combine the numerators over the common denominator.

$8^{\frac{x \log (x^{2} - 3) + \log ({(x^{2} - 3)}^{2})}{\log (8)}}$