Difference between revisions of "1956 AHSME Problems/Problem 8"

(Problem 8)
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<math>\textbf{(A)}\ - 4 \qquad\textbf{(B)}\ - 3 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 8 </math>
 
<math>\textbf{(A)}\ - 4 \qquad\textbf{(B)}\ - 3 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 8 </math>
 +
<math>7^1 cdot 5^3 cdot 3^1 cdot 2^4</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 22:23, 2 June 2022

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

See Also

1956 AHSME (ProblemsAnswer KeyResources)
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
All AHSME Problems and Solutions

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