Difference between revisions of "1968 AHSME Problems/Problem 2"

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== Solution ==
 
== Solution ==
<math>\fbox{}</math>
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<math>\fbox{B}</math>
  
 
== See also ==
 
== See also ==

Revision as of 02:25, 29 September 2014

Problem

The real value of $x$ such that $64^{x-1}$ divided by $4^{x-1}$ equals $256^{2x}$ is:

$\text{(A) } -\frac{2}{3}\quad\text{(B) } -\frac{1}{3}\quad\text{(C) } 0\quad\text{(D) } \frac{1}{4}\quad\text{(E) } \frac{3}{8}$



Solution

$\fbox{B}$

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 2
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All AHSME Problems and Solutions

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