Find the Exact Value log base 2 of 3/25+5 log base 2 of 5+ log base 2 of 27- log base 2 of 125/12- log base 2 of 243

$\log_{2} (\frac{3}{25}) + 5 \log_{2} (5) + \log_{2} (27) - \log_{2} (\frac{125}{12}) - \log_{2} (243)$

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

$5 \log_{2} (5) + \log_{2} (\frac{3}{25} \cdot 27) - \log_{2} (\frac{125}{12}) - \log_{2} (243)$

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

$5 \log_{2} (5) + \log_{2} (\frac{\frac{3}{25} \cdot 27}{\frac{125}{12}}) - \log_{2} (243)$

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

$5 \log_{2} (5) + \log_{2} (\frac{\frac{\frac{3}{25} \cdot 27}{\frac{125}{12}}}{243})$

Simplify each term.

Simplify $5 \log_{2} (5)$ by moving $5$ inside the logarithm.

$\log_{2} (5^{5}) + \log_{2} (\frac{\frac{\frac{3}{25} \cdot 27}{\frac{125}{12}}}{243})$

Raise $5$ to the power of $5$ .

$\log_{2} (3125) + \log_{2} (\frac{\frac{\frac{3}{25} \cdot 27}{\frac{125}{12}}}{243})$

Multiply the numerator by the reciprocal of the denominator.

$\log_{2} (3125) + \log_{2} (\frac{\frac{3}{25} \cdot 27}{\frac{125}{12}} \cdot \frac{1}{243})$

Cancel the common factor of $27$ .

Factor $27$ out of $\frac{3}{25} \cdot 27$ .

$\log_{2} (3125) + \log_{2} (\frac{27 \cdot \frac{3}{25}}{\frac{125}{12}} \cdot \frac{1}{243})$

Factor $27$ out of $243$ .

$\log_{2} (3125) + \log_{2} (\frac{27 \cdot \frac{3}{25}}{\frac{125}{12}} \cdot \frac{1}{27 (9)})$

Cancel the common factor.

$\log_{2} (3125) + \log_{2} (\frac{27 \cdot \frac{3}{25}}{\frac{125}{12}} \cdot \frac{1}{27 \cdot 9})$

Rewrite the expression.

$\log_{2} (3125) + \log_{2} (\frac{\frac{3}{25}}{\frac{125}{12}} \cdot \frac{1}{9})$

Multiply $\frac{\frac{3}{25}}{\frac{125}{12}}$ by $\frac{1}{9}$ .

$\log_{2} (3125) + \log_{2} (\frac{\frac{3}{25}}{\frac{125}{12} \cdot 9})$

Combine $\frac{125}{12}$ and $9$ .

$\log_{2} (3125) + \log_{2} (\frac{\frac{3}{25}}{\frac{125 \cdot 9}{12}})$

Multiply $125$ by $9$ .

$\log_{2} (3125) + \log_{2} (\frac{\frac{3}{25}}{\frac{1125}{12}})$

Cancel the common factor of $1125$ and $12$ .

Factor $3$ out of $1125$ .

$\log_{2} (3125) + \log_{2} (\frac{\frac{3}{25}}{\frac{3 (375)}{12}})$

Cancel the common factors.

Factor $3$ out of $12$ .

$\log_{2} (3125) + \log_{2} (\frac{\frac{3}{25}}{\frac{3 \cdot 375}{3 \cdot 4}})$

Cancel the common factor.

$\log_{2} (3125) + \log_{2} (\frac{\frac{3}{25}}{\frac{3 \cdot 375}{3 \cdot 4}})$

Rewrite the expression.

$\log_{2} (3125) + \log_{2} (\frac{\frac{3}{25}}{\frac{375}{4}})$

Multiply the numerator by the reciprocal of the denominator.

$\log_{2} (3125) + \log_{2} (\frac{3}{25} \cdot \frac{4}{375})$

Cancel the common factor of $3$ .

Factor $3$ out of $375$ .

$\log_{2} (3125) + \log_{2} (\frac{3}{25} \cdot \frac{4}{3 (125)})$

Cancel the common factor.

$\log_{2} (3125) + \log_{2} (\frac{3}{25} \cdot \frac{4}{3 \cdot 125})$

Rewrite the expression.

$\log_{2} (3125) + \log_{2} (\frac{1}{25} \cdot \frac{4}{125})$

Multiply $\frac{1}{25}$ by $\frac{4}{125}$ .

$\log_{2} (3125) + \log_{2} (\frac{4}{25 \cdot 125})$

Multiply $25$ by $125$ .

$\log_{2} (3125) + \log_{2} (\frac{4}{3125})$

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

$\log_{2} (3125 (\frac{4}{3125}))$

Cancel the common factor of $3125$ .

Cancel the common factor.

$\log_{2} (3125 (\frac{4}{3125}))$

Rewrite the expression.

$\log_{2} (4)$

Logarithm base $2$ of $4$ is $2$ .

$2$

Find the Exact Value log base 2 of 3/25+5 log base 2 of 5+ log base 2 of 27- log base 2 of 125/12- log base 2 of 243

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