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

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(Problem 8)
 
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== Problem 8==
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If <math>8\cdot2^x = 5^{y + 8}</math>, then when <math>y = - 8,x = </math>
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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>
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== Solution ==
 
== Solution ==
  
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This simply gives that <math>x=-3</math>.
 
This simply gives that <math>x=-3</math>.
 
Therefore, the answer is <math>\fbox{(B) -3}</math>.
 
Therefore, the answer is <math>\fbox{(B) -3}</math>.
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==See Also==
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{{AHSME box|year=1956|num-b=7|num-a=9}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:25, 2 June 2022

Problem 8

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

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

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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