Difference between revisions of "1956 AHSME Problems/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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Dividing by 8 gives | Dividing by 8 gives | ||
<cmath>2^{x}=\frac{1}{8}</cmath> | <cmath>2^{x}=\frac{1}{8}</cmath> | ||
| − | This simply gives that <math>x=-3</math> | + | This simply gives that <math>x=-3</math>. |
| − | Therefore, the answer is <math>\fbox{( | + | 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]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:25, 2 June 2022
Problem 8
If 
, then when 
Solution
Simple substitution yields
![\[8 \cdot 2^{x} = 5^{0}\]](http://latex.artofproblemsolving.com/d/6/a/d6af6b14becb25e736b1d5508e3b4439c16d4091.png)
Reducing the equation gives
![\[8 \cdot 2^{x} = 1\]](http://latex.artofproblemsolving.com/0/0/2/0029b0b2a856bad4e423489d186011cd93e1c355.png)
Dividing by 8 gives
![\[2^{x}=\frac{1}{8}\]](http://latex.artofproblemsolving.com/8/a/4/8a40ad1d77235bf0f7ca675e0ebab1452efc7fde.png)
This simply gives that 
.
Therefore, the answer is 
.
See Also
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
