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x is a positive, odd integer.

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Intern
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Re: x is a positive, odd integer. [#permalink] New post 09 Jul 2018, 17:15
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So the way I approached this was by doing a guess and check method.

The values for 'x' used: 1, 2, -1. -2

x = 1 --> (-3)^1 = -3 & -2^2(1) = -4 ---> -3 > -4
x = 2 --> (-3)^2 = 9 & -2^2(2) = -16 ---> 9 > -16
x = -1 --> (-3)^(-1) = -1/3 & -2^2(-1) = -1/4 ---> -1/3 > -1/4
x = -2 --> (-3)^(-2) = 1/9 & -2^2(-2) = -1/16 ---> 1/9 > -1/16

Based on the calculations we completed, it is absolutly clear that the answer is A.
Intern
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Kudos [?]: 2 [0], given: 2

Re: x is a positive, odd integer. [#permalink] New post 15 Aug 2018, 20:15
There is a major flaw in the Question and its solution, because my education tells me and so does mathsfirst.massey.ac.nz/Algebra/OrderOfOp/orderAlg.htm look at the last example of Implied Brackets, one must solve the Power, if it is an expression, before he/she raise the number to that power.

I think this is an erroneous use of PEMDAS here.
Re: x is a positive, odd integer.   [#permalink] 15 Aug 2018, 20:15
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