Precalculus Examples

$\log_{5.656} (2)$

Rewrite $\log_{5.656} (2)$ using the change of base formula.

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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 $0$ .

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

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

$\frac{\log (2)}{\log (5.656)}$

Log base $10$ of $5.656$ is approximately $0.7525094$ .

$\frac{\log (2)}{0.7525094}$

Log base $10$ of $2$ is approximately $0.30102999$ .

$\frac{0.30102999}{0.7525094}$

Divide $0.30102999$ by $0.7525094$ .

$0.40003486$