# Difference between revisions of "1993 AHSME Problems/Problem 10"

## Problem

Let $r$ be the number that results when both the base and the exponent of $a^b$ are tripled, where $a,b>0$. If $r$ equals the product of $a^b$ and $x^b$ where $x>0$, then $x=$

$\text{(A) } 3\quad \text{(B) } 3a^2\quad \text{(C) } 27a^2\quad \text{(D) } 2a^{3b}\quad \text{(E) } 3a^{2b}$

## Solution

We have $r=(3a)^{3b}$

From this we have the equation $(3a)^{3b}=a^bx^b$

Raising both sides to the $\frac{1}{b}$ power we get that $27a^3=ax$ or $x=27a^2$

$\fbox{C}$