Difference between revisions of "1958 AHSME Problems/Problem 20"

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== Problem ==
 
== Problem ==
If <math> 4^x \minus{} 4^{x \minus{} 1} \equal{} 24</math>, then <math> (2x)^x</math>  equals:
+
If <math> 4^x - 4^{x - 1} = 24</math>, then <math> (2x)^x</math>  equals:
  
 
<math> \textbf{(A)}\ 5\sqrt{5}\qquad  
 
<math> \textbf{(A)}\ 5\sqrt{5}\qquad  

Latest revision as of 23:19, 13 March 2015

Problem

If $4^x - 4^{x - 1} = 24$, then $(2x)^x$ equals:

$\textbf{(A)}\ 5\sqrt{5}\qquad  \textbf{(B)}\ \sqrt{5}\qquad  \textbf{(C)}\ 25\sqrt{5}\qquad  \textbf{(D)}\ 125\qquad  \textbf{(E)}\ 25$

Solution

$\fbox{}$

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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