Precalculus Examples

$\log_{2} (6) - \log_{2} (15) + \log_{2} (80)$

Step 1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{2} (\frac{6}{15}) + \log_{2} (80)$

Step 2

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

$\log_{2} (\frac{6}{15} \cdot 80)$

Step 3

Cancel the common factor of $5$ .

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Step 3.1

Factor $5$ out of $15$ .

$\log_{2} (\frac{6}{5 (3)} \cdot 80)$

Step 3.2

Factor $5$ out of $80$ .

$\log_{2} (\frac{6}{5 \cdot 3} \cdot (5 \cdot 16))$

Step 3.3

Cancel the common factor.

$\log_{2} (\frac{6}{5 \cdot 3} \cdot (5 \cdot 16))$

Step 3.4

Rewrite the expression.

$\log_{2} (\frac{6}{3} \cdot 16)$

Step 4

Combine $\frac{6}{3}$ and $16$ .

$\log_{2} (\frac{6 \cdot 16}{3})$

Step 5

Simplify the expression.

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Step 5.1

Multiply $6$ by $16$ .

$\log_{2} (\frac{96}{3})$

Step 5.2

Logarithm base $2$ of $\frac{96}{3}$ is $5$ .

$5$