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: \begin{align*} \frac{64^{x-1}}{4^{x-1}}=256^{2x} \\ \frac{(4^3)^{x-1}}{4^{x-1}}=(4^4)^{2x} \\ \frac{4^{3x-3}}{4^{x-1}}=4^{8x} \\ 4^{2x-2}=4^{8x} \\ 2x-2=8x \\ x-1=4x \\ 3x=-1 \\ x=\frac{-1}{3} \\ \end{align*}

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

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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1968 AHSME Problems/Problem 2

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