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

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

Latest revision as of 22:00, 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
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

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