Evaluate log base 7 of 7^3+ log base 7 of 7^8

$\log_{7} (7^{3}) + \log_{7} (7^{8})$

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

$\log_{7} (7^{3} \cdot 7^{8})$

Multiply $7^{3}$ by $7^{8}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\log_{7} (7^{3 + 8})$

Add $3$ and $8$ .

$\log_{7} (7^{11})$

Use logarithm rules to move $11$ out of the exponent.

$11 \log_{7} (7)$

Logarithm base $7$ of $7$ is $1$ .

$11 \cdot 1$

Multiply $11$ by $1$ .

$11$