Find the Exact Value log base 2 of 6- log base 2 of 15+ log base 2 of 20

$\log_{2} (6) - \log_{2} (15) + \log_{2} (20)$

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

$\log_{2} (\frac{6}{15}) + \log_{2} (20)$

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

$\log_{2} (\frac{6}{15} \cdot 20)$

Cancel the common factor of $5$ .

Factor $5$ out of $15$ .

$\log_{2} (\frac{6}{5 (3)} \cdot 20)$

Factor $5$ out of $20$ .

$\log_{2} (\frac{6}{5 \cdot 3} \cdot (5 \cdot 4))$

Cancel the common factor.

$\log_{2} (\frac{6}{5 \cdot 3} \cdot (5 \cdot 4))$

Rewrite the expression.

$\log_{2} (\frac{6}{3} \cdot 4)$

Combine $\frac{6}{3}$ and $4$ .

$\log_{2} (\frac{6 \cdot 4}{3})$

Simplify the expression.

Multiply $6$ by $4$ .

$\log_{2} (\frac{24}{3})$

Logarithm base $2$ of $\frac{24}{3}$ is $3$ .

$3$