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 $x=27a^2$

$\fbox{C}$

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1993 AHSME Problems/Problem 10

Problem

Solution

See also