Precalculus Examples

$\log_{20} (100) (\log (5) + \log (4))$

Step 1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{20} (100) \log (5 \cdot 4)$

Step 2

Multiply $5$ by $4$ .

$\log_{20} (100) \log (20)$

Step 3

Rewrite $\log_{20} (100)$ 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 $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log (20)$

Step 3.2

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

$\frac{\log (100)}{\log (20)} \log (20)$

Step 4

Cancel the common factor of $\log (20)$ .

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Step 4.1

Logarithm base $10$ of $100$ is $2$ .

$\frac{2}{\log (20)} \log (20)$

Step 4.2

Cancel the common factor.

$\frac{2}{\log (20)} \log (20)$

Step 4.3

Rewrite the expression.

$2$