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

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== Solution ==
 
== Solution ==
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We have <math>r=(3a)^{3b}</math>
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From this we have the equation <math>(3a)^{3b}=a^bx^b</math>
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Raising both sides to the <math>\frac{1}{b}</math> power we get that <math>27a^3=ax</math> or  <math>x=27a^2</math>
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<math>\fbox{C}</math>
 
<math>\fbox{C}</math>
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1993|num-b=1|num-a=2}}   
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{{AHSME box|year=1993|num-b=9|num-a=11}}   
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:00, 27 May 2021

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 $x=27a^2$

$\fbox{C}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AHSME Problems and Solutions

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