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Difference between revisions of "1971 AHSME Problems/Problem 7"
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From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule.\  
 
From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule.\  
 
\\*Now\ we\ can\ see\ this\ simplies\ to\ \frac{-x}{2}\
 
\\*Now\ we\ can\ see\ this\ simplies\ to\ \frac{-x}{2}\
−
\\*Looking\ at\ \frac{x}{2} we\ can\ clearly\ see\ that\ \frac{-x}{2} is\ equal\ to\ -2^{-(2k+1)}\
+
\\*Looking\ at\ \frac{x}{2} we\ can\ clearly\ see\ that\ -(\frac{x}{2}) is\ equal\ to\ -2^{-(2k+1)}\
 
\\*Thus\ our\ answer\ is\ c</math>
 
\\*Thus\ our\ answer\ is\ c</math>

Revision as of 19:57, 29 April 2020

$Let\ x\ equal\ 2^{-2k}\ \\* From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule.\ \\*Now\ we\ can\ see\ this\ simplies\ to\ \frac{-x}{2}\ \\*Looking\ at\ \frac{x}{2} we\ can\ clearly\ see\ that\ -(\frac{x}{2}) is\ equal\ to\ -2^{-(2k+1)}\ \\*Thus\ our\ answer\ is\ c$

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