Answer
$x=1$ and $x=-8$
Work Step by Step
If $b\gt0$ and $b\ne1$, then $y=log_{b}x$ means $x=b^{y}$ for every $x\gt0$ and every real number $y$.
Therefore, $log_{8}(x^{2}+7x)=1$ means $8^{1}=8=x^{2}+7x$.
Subtract 8 from both sides of the equation to get all terms on one side.
$x^{2}+7x-8=0$
We know that 1 and 8 are factors of 8 (and the sum of -1 and 8 is 7, which is the coefficient attached to the middle term).
Therefore, we can factor this equation into $(x-1)(x+8)=0$. After setting both terms equal to 0, we know that $x=1$ and $x=-8$.
