1956 AHSME Problems/Problem 8

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Problem 8

If $8\cdot2^x = 5^{y + 8}$, then when $y = - 8,x =$

$\textbf{(A)}\ - 4 \qquad\textbf{(B)}\ - 3 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 8$

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 $\fbox{(B) -3}$.

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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