Write as a Single Logarithm 2( log base 3 of 8+ log base 3 of z)- log base 3 of 3^4-7^2

$2 (\log_{3} (8) + \log_{3} (z)) - \log_{3} (3^{4} - 7^{2})$

Step 1

Simplify each term.

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

$2 \log_{3} (8 z) - \log_{3} (3^{4} - 7^{2})$

Simplify $2 \log_{3} (8 z)$ by moving $2$ inside the logarithm.

$\log_{3} ({(8 z)}^{2}) - \log_{3} (3^{4} - 7^{2})$

Apply the product rule to $8 z$ .

$\log_{3} (8^{2} z^{2}) - \log_{3} (3^{4} - 7^{2})$

Raise $8$ to the power of $2$ .

$\log_{3} (64 z^{2}) - \log_{3} (3^{4} - 7^{2})$

Simplify each term.

Raise $3$ to the power of $4$ .

$\log_{3} (64 z^{2}) - \log_{3} (81 - 7^{2})$

Raise $7$ to the power of $2$ .

$\log_{3} (64 z^{2}) - \log_{3} (81 - 1 \cdot 49)$

Multiply $- 1$ by $49$ .

$\log_{3} (64 z^{2}) - \log_{3} (81 - 49)$

Subtract $49$ from $81$ .

$\log_{3} (64 z^{2}) - \log_{3} (32)$

Step 2

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

$\log_{3} (\frac{64 z^{2}}{32})$

Step 3

Cancel the common factor of $64$ and $32$ .

Factor $32$ out of $64 z^{2}$ .

$\log_{3} (\frac{32 (2 z^{2})}{32})$

Cancel the common factors.

Factor $32$ out of $32$ .

$\log_{3} (\frac{32 (2 z^{2})}{32 (1)})$

Cancel the common factor.

$\log_{3} (\frac{32 (2 z^{2})}{32 \cdot 1})$

Rewrite the expression.

$\log_{3} (\frac{2 z^{2}}{1})$

Divide $2 z^{2}$ by $1$ .

$\log_{3} (2 z^{2})$