1956 AHSME Problems/Problem 8

Problem 8

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

$\textbf{(A)}\ - 4 \qquad\textbf{(B)}\ - 3 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 8$ $7^1 cdot 5^3 cdot 3^1 cdot 2^4$

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

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1956 AHSME Problems/Problem 8

Problem 8

Solution

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