Precalculus Examples

$(2 \log_{b} (7)) (\log_{b} (8))$

Step 1

Simplify $2 \log_{b} (7)$ by moving $2$ inside the logarithm.

$\log_{b} (7^{2}) \log_{b} (8)$

Step 2

Raise $7$ to the power of $2$ .

$\log_{b} (49) \log_{b} (8)$

Step 3

Rewrite $\log_{b} (49)$ using the change of base formula.

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Step 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 $0$ .

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

Step 3.2

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

$\frac{\log (49)}{\log (b)} \log_{b} (8)$

Step 4

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

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Step 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 $0$ .

$\frac{\log (49)}{\log (b)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 4.2

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

$\frac{\log (49)}{\log (b)} \cdot \frac{\log (8)}{\log (b)}$

Step 5

Multiply $\frac{\log (49)}{\log (b)} \cdot \frac{\log (8)}{\log (b)}$ .

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Step 5.1

Multiply $\frac{\log (49)}{\log (b)}$ by $\frac{\log (8)}{\log (b)}$ .

$\frac{\log (49) \log (8)}{\log (b) \log (b)}$

Step 5.2

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

$\frac{\log (49) \log (8)}{\log^{1} (b) \log (b)}$

Step 5.3

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

$\frac{\log (49) \log (8)}{\log^{1} (b) \log^{1} (b)}$

Step 5.4

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

$\frac{\log (49) \log (8)}{{\log (b)}^{1 + 1}}$

Step 5.5

Add $1$ and $1$ .

$\frac{\log (49) \log (8)}{\log^{2} (b)}$