Find the Exact Value log base 3 of 54- log base 3 of 8/15+ log base 3 of 4/5

$\log_{3} (54) - \log_{3} (\frac{8}{15}) + \log_{3} (\frac{4}{5})$

Step 1

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

$\log_{3} (\frac{54}{\frac{8}{15}}) + \log_{3} (\frac{4}{5})$

Step 2

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

$\log_{3} (\frac{54}{\frac{8}{15}} \cdot \frac{4}{5})$

Step 3

Multiply the numerator by the reciprocal of the denominator.

$\log_{3} (54 (\frac{15}{8}) \frac{4}{5})$

Step 4

Cancel the common factor of $2$ .

Factor $2$ out of $54$ .

$\log_{3} (2 (27) \frac{15}{8} \frac{4}{5})$

Factor $2$ out of $8$ .

$\log_{3} (2 \cdot 27 \frac{15}{2 \cdot 4} \frac{4}{5})$

Cancel the common factor.

$\log_{3} (2 \cdot 27 \frac{15}{2 \cdot 4} \frac{4}{5})$

Rewrite the expression.

$\log_{3} (27 (\frac{15}{4}) \frac{4}{5})$

Step 5

Combine $27$ and $\frac{15}{4}$ .

$\log_{3} (\frac{27 \cdot 15}{4} \cdot \frac{4}{5})$

Step 6

Multiply $27$ by $15$ .

$\log_{3} (\frac{405}{4} \cdot \frac{4}{5})$

Step 7

Cancel the common factor of $5$ .

Factor $5$ out of $405$ .

$\log_{3} (\frac{5 (81)}{4} \cdot \frac{4}{5})$

Cancel the common factor.

$\log_{3} (\frac{5 \cdot 81}{4} \cdot \frac{4}{5})$

Rewrite the expression.

$\log_{3} (\frac{81}{4} \cdot 4)$

Step 8

Cancel the common factor of $4$ .

Cancel the common factor.

$\log_{3} (\frac{81}{4} \cdot 4)$

Rewrite the expression.

$\log_{3} (81)$

Step 9

Logarithm base $3$ of $81$ is $4$ .

$4$