1968 AHSME Problems/Problem 2

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

Because $64=4^3$ and $256=4^4$, we can change all of the numbers in the equation to exponents with base $4$ and solve the equation: 64x14x1=2562x(43)x14x1=(44)2x43x34x1=48x42x2=48x2x2=8xx1=4x3x=1x=13

Thus, our desired answer is $\fbox{(B) -1/3}$.

See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 31 32 33 34 35
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