smc

Understand that ${\log }_{10}\left(x\right)=y$ can be expressed as $x=10^y$. By applying this fact to ${\log }_{10}\left(a^2-15a\right)=2$, we get $a^2-15a=10^2$ We can use factoring methods to bring us to $(a-20)(a+5)=0$, which, as each binomial produces one integer solution, gives us a solution set of $\boxed{\text{two integers }}$.