1965 AHSME Problems/Problem 6

Problem 6

If $10^{\log_{10}9} = 8x + 5$ then $x$ equals:

$\textbf{(A)}\ 0 \qquad  \textbf{(B) }\ \frac {1}{2} \qquad  \textbf{(C) }\ \frac {5}{8} \qquad  \textbf{(D) }\ \frac{9}{8}\qquad \textbf{(E) }\ \frac{2\log_{10}3-5}{8}$

Solution

Notice that $10^{\log_{10} 9} = 9$. Therefore, the condition we are looking for is $9=8x+5$, or $x=\frac{1}{2}$. $\text{So the answer is }  \boxed{\textbf{(B)}}$

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