Write as a Single Logarithm log base 3 of 12+ log base 3 of 4.5- log base 3 of 6

$\log_{3} (12) + \log_{3} (4.5) - \log_{3} (6)$

Step 1

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

$\log_{3} (12 \cdot 4.5) - \log_{3} (6)$

Step 2

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

$\log_{3} (\frac{12 \cdot 4.5}{6})$

Step 3

Cancel the common factor of $12$ and $6$ .

Factor $6$ out of $12 \cdot 4.5$ .

$\log_{3} (\frac{6 (2 \cdot 4.5)}{6})$

Cancel the common factors.

Factor $6$ out of $6$ .

$\log_{3} (\frac{6 (2 \cdot 4.5)}{6 (1)})$

Cancel the common factor.

$\log_{3} (\frac{6 (2 \cdot 4.5)}{6 \cdot 1})$

Rewrite the expression.

$\log_{3} (\frac{2 \cdot 4.5}{1})$

Divide $2 \cdot 4.5$ by $1$ .

$\log_{3} (2 \cdot 4.5)$

Step 4

Multiply $2$ by $4.5$ .

$\log_{3} (9)$

Step 5

Log base $3$ of $9$ is approximately $2$ .

$2$