# Difference between revisions of "1951 AHSME Problems/Problem 22"

## Problem

The values of $a$ in the equation: $\log_{10}(a^2 - 15a) = 2$ are:

$\textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\textbf{(B)}\ 20, - 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}$ $\textbf{(D)}\ \pm20 \qquad\textbf{(E)}\ \text{none of these}$

## Solution

Putting into exponential form, we get that $10^2=a^2-15a\Rightarrow a^2-15a-100=0$

Now we use the quadratic formula to solve for $a$, and we get $a=\frac{15\pm\sqrt{625}}{2}\implies a=\boxed{\textbf{(B)}\ 20, - 5}$