Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎See also)
(→‎Solution)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
 +
We have
 +
<math>r=(3a)^{3b}</math>
 +
\\
 +
From this we have the equation
 +
<math>(3a)^{3b}=a^bx^b</math>
 +
Raising both sides to the <math>\frac{1}{b}</math> power we get that
 +
<math>27a^3=ax</math>
 +
<math>x=27a^2</math>
 
<math>\fbox{C}</math>
 
<math>\fbox{C}</math>
  

Revision as of 15:47, 8 August 2016

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

See also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_10&oldid=79902"