If $10^{\log_{10}9} = 8x + 5$ then $x$ equals: 
$\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ \frac {1}{2} \qquad  \textbf{(C) }\ \frac {5}{8} \qquad  \textbf{(D) }\ \frac{9}{8}\qquad \textbf{(E) }\ \frac{2\log_{10}3-5}{8}$
Notice that $10^{\log_{10} 9} = 9$. Therefore, the condition we are looking for is $9=8x+5$, or $x=\frac{1}{2}$. $\text{So the answer is }  \boxed{\textbf{(B)}}$