Answer
$\log_b1$ is $0$, so $\log_b8 + \log_b1 = \log_b8 + 0 = \log_b8 $.
Work Step by Step
As long as the value of $b$ is greater than 0 and not 1, $\log_b1$ is always equal to $0$. This is because $b^x=1$, where b is greater than 0 and not 1, is only true when $x=0$.
