1956 AHSME Problems/Problem 8

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Solution

Simple substitution yields \[8 \cdot 2^{x} = 5^{0}\] Reducing the equation gives \[8 \cdot 2^{x} = 1\] Dividing by 8 gives

\[2^{x}=\frac{1]{8}\] (Error compiling LaTeX. Unknown error_msg)

This simply gives that $x=-3$ Therefore, the answer is $\fbox{(b) -30}$