# 1959 AHSME Problems/Problem 23

## Problem 23

The set of solutions of the equation $\log_{10}\left( a^2-15a\right)=2$ consists of $\textbf{(A)}\ \text{two integers } \qquad\textbf{(B)}\ \text{one integer and one fraction}\qquad \textbf{(C)}\ \text{two irrational numbers }\qquad\textbf{(D)}\ \text{two non-real numbers} \qquad\textbf{(E)}\ \text{no numbers, that is, the empty set}$

## Solution

Understand that $\log_{10}\left(x\right)=y$ can be expressed as $x = y^2$.

By applying that same logic 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 set of $\textbf{(A)}$ two integers in the solution set.

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