1956 AHSME Problems/Problem 8

Solution

Simple substitution yields $8 \cdot 2^{x} = 5^{0}$ Reducing the equation gives $8 \cdot 2^{x} = 1$ Dividing by 8 gives $2^{x}=\frac{1}{8}$ This simply gives that $x=-3$ Therefore, the answer is $\fbox{(B) -3}$

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1956 AHSME Problems/Problem 8

Solution