Evaluate log base 3 of 12- log base 3 of 28+ log base 3 of 63

$\log_{3} (12) - \log_{3} (28) + \log_{3} (63)$

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

$\log_{3} (\frac{12}{28}) + \log_{3} (63)$

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

$\log_{3} (\frac{12}{28} \cdot 63)$

Cancel the common factor of $7$ .

Factor $7$ out of $28$ .

$\log_{3} (\frac{12}{7 (4)} \cdot 63)$

Factor $7$ out of $63$ .

$\log_{3} (\frac{12}{7 \cdot 4} \cdot (7 \cdot 9))$

Cancel the common factor.

$\log_{3} (\frac{12}{7 \cdot 4} \cdot (7 \cdot 9))$

Rewrite the expression.

$\log_{3} (\frac{12}{4} \cdot 9)$

Combine $\frac{12}{4}$ and $9$ .

$\log_{3} (\frac{12 \cdot 9}{4})$

Simplify the expression.

Multiply $12$ by $9$ .

$\log_{3} (\frac{108}{4})$

Logarithm base $3$ of $\frac{108}{4}$ is $3$ .

$3$