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 $27a^3=ax$ $x=27a^2$ $\fbox{C}$

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1993 AHSME Problems/Problem 10

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