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by mysseo » Mon Dec 19, 2011 6:02 pm
Each of the following equations has at least one solution EXCEPT
A. -2^n = (-2)^-n
B. 2^-n = (-2)^n
C. 2^n = (-2)^-n
D. (-2)^n = -2^n
E. (-2)^-n = -2^-n


It was on Manhattan CAT. Parenthesis makes me confused.
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by GmatMathPro » Mon Dec 19, 2011 6:12 pm
mysseo wrote:Each of the following equations has at least one solution EXCEPT
A. -2^n = (-2)^-n
B. 2^-n = (-2)^n
C. 2^n = (-2)^-n
D. (-2)^n = -2^n
E. (-2)^-n = -2^-n


It was on Manhattan CAT. I am not sure how to solve it.
I would start by plugging simple values for n into the equations. Anything (except zero) to the zero power is one, so that would be the simplest place to start. Note that n=0 is a solution for B & C. Now try n=1. n=1 is a solution to D & E. Therefore, the correct answer is A.
Last edited by GmatMathPro on Mon Dec 19, 2011 6:36 pm, edited 1 time in total.
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by mysseo » Mon Dec 19, 2011 6:32 pm
Why isn't A's solution n=0? -2^0=0, (-2)^-0=0.
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by GmatMathPro » Mon Dec 19, 2011 6:35 pm
mysseo wrote:Why isn't A's solution n=0? -2^0=0, (-2)^-0=0.
Remember that anything (except zero) to the zero power is one, not zero, so -2^0=-1 and (-2)^-0=1.
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by mysseo » Mon Dec 19, 2011 6:43 pm
I got it. Thank you so much for your help.
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