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March 4th, 2014, 04:31 AM   #1
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Joined: Dec 2013

Posts: 22
Thanks: 2

Integer Exponent Problem

Hello,

I've done this problem a number of times and can't seem to come to the correct answer. Here is the problem:

(2^-2+2^-1+2^0)^-2

According to WolframAlpha the solution is 16/49 but I always get 16/21. There must be a property or a rule that is missing somewhere. I uploaded an image of my work for the problem for reference.

Any insight as to how to solve the problem to get the correct solution would be appreciated!

Thanks,

~Kiphyl
Attached Images
 theWork.jpg (36.4 KB, 66 views)

 March 4th, 2014, 05:00 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Integer Exponent Problem Your error is in the fourth line where $\left(\frac{1}{2^2}+ \frac{1}{2}+ \frac{2}{2}\right)^2$ has become $\frac{1}{2^4}+ \frac{1}{2^2}+ \frac{2^2}{2^2}$. That is the fairly common error of thinking that $(a+ b+ c)^2= a^2+ b^2+ c^2$. Instead do the addition before squaring: $\frac{1}{2^2}+ \frac{1}{2}+ \frac{2}{2}= \frac{1}{4}+ \frac{2}{4}+ \frac{4}{4}= \frac{7}{4}$ and squaring now gives $\frac{49}{16}$. 1 over that is, of course, $\frac{16}{49}$.
 March 4th, 2014, 05:55 AM #3 Newbie   Joined: Dec 2013 Posts: 22 Thanks: 2 Re: Integer Exponent Problem Thank you so much!!!
 March 4th, 2014, 09:24 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,474 Thanks: 2039 $\frac{1}{$$2^{-2}\,+\,2^{-1}\,+\,2^0$$^2}\,=\,\frac{$$2^2$$^2}{$$2^2\(2^{-2}\,+\,2^{-1}\,+\,2^0$$\)^2}\,=\,\frac{4^2}{$$2^0\,+\,2^1\,+\ ,2^2$$^2}\,=\,\frac{16}{(1\,+\,2\,+\,4)^2}\,=\,\fr ac{16}{49}$

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