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1956 AHSME Problems/Problem 8

Revision as of 14:43, 31 August 2019 by Mr.peppa (talk | contribs) (→‎Solution)

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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