Difference between revisions of "1971 AHSME Problems/Problem 7"
Logical231 (talk | contribs) (Created page with "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...") |
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| − | Let\ x\ equal\ 2^{-2k} | + | <cmath>Let\ x\ equal\ 2^{-2k} |
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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 | + | Thus\ our\ answer\ is\ c</cmath> |
Revision as of 20:22, 29 April 2020
![\[Let\ x\ equal\ 2^{-2k} \ From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule\ Now\ we\ can\ see\ this\ simplies\ to\ \frac{-x}{2}\ Looking\ at\ \frac{x}{2} we\ can\ clearly\ see\ that\ \frac{-x}{2} is\ equal\ to\ -2^{-(2k+1)}\ Thus\ our\ answer\ is\ c\]](http://latex.artofproblemsolving.com/2/5/a/25ab732e8e93e0a0faea435af73cecc1dc9a7e36.png)
