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# Each of the following equations has at least one solution

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Each of the following has at least one solution EXCEPT [#permalink]

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23 Aug 2012, 03:45
OA has to be A because
Equation 1 simplifies to (2)^n (2)^n (-1)^n= -1 has no solution for any value of n
Rest of options have at least 1 solution
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Re: Each of the following equations has at least one solution [#permalink]

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08 Oct 2012, 04:00
if -2^n means -(2^n), the answer is A. Otherwise I find 0 to satisfy all equations.
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Re: Each of the following equations has at least one solution [#permalink]

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08 Oct 2012, 04:26
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14 Oct 2012, 07:05
Bunuel wrote:
gurpreetsingh wrote:
all seems to have n=0 as solution....? whats the OA

$$n=0$$ is not a solution of the equation $$-2^n = (-2)^{-n}$$ (in fact this equation has no solution):

$$-2^n=-(2^n)=-(2^{0})=-1$$ but $$(-2)^{-n}=(-2)^{0}=1$$.

I would like to double check why we say that n=0 could be a solution in case of $$(-2)^{-n}$$
as $$(-2)^{-n} = (-2)^{1/n}$$ and then we can not divide by zero?

Nik

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14 Oct 2012, 22:12
NikRu wrote:
I would like to double check why we say that n=0 could be a solution in case of $$(-2)^{-n}$$
as $$(-2)^{-n} = (-2)^{1/n}$$ and then we can not divide by zero?

Nik

$$(-2)^{-n} = 1/(-2)^n$$ not $$(-2)^{1/n}$$
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18465 [0], given: 237 Senior Manager Joined: 03 Dec 2012 Posts: 328 Kudos [?]: 198 [0], given: 291 Re: Each of the following equations has at least one solution [#permalink] ### Show Tags 26 Oct 2013, 03:40 Can somebody please explain if (–2)^n = 1 or -1 (if n=0) Kudos [?]: 198 [0], given: 291 Math Expert Joined: 02 Sep 2009 Posts: 43320 Kudos [?]: 139373 [1], given: 12787 Re: Each of the following equations has at least one solution [#permalink] ### Show Tags 26 Oct 2013, 03:44 1 This post received KUDOS Expert's post mohnish104 wrote: Can somebody please explain if (–2)^n = 1 or -1 (if n=0) Check here: each-of-the-following-equations-has-at-least-one-solution-94119.html#p738365 _________________ Kudos [?]: 139373 [1], given: 12787 Current Student Joined: 06 Sep 2013 Posts: 1957 Kudos [?]: 770 [0], given: 355 Concentration: Finance Re: Exponents [#permalink] ### Show Tags 15 Jan 2014, 08:57 nverma wrote: marcusaurelius wrote: Each of the following equations has at least one solution EXCEPT –2^n = (–2)^-n 2^-n = (–2)^n 2^n = (–2)^-n (–2)^n = –2^n (–2)^-n = –2^-n IMHO A a) –2^n = (–2)^-n –2^n = 1/(–2)^n –2^n * (–2)^n = 1, Keep it. Let's solve the other options..!! b) 2^-n = (–2)^n 1/2^n = (–2)^n 1 = (–2)^n * (2^n) For n=0, L.H.S = R.H.S c) 2^n = (–2)^-n 2^n = 1/ (–2)^n (2^n) * (–2)^n = 1 For n=0, L.H.S = R.H.S d) (–2)^n = –2^n (–2)^n / –2^n = 1 For n=1, L.H.S = R.H.S e) (–2)^-n = –2^-n 1/ (–2)^n = 1/–2^n For n=1, L.H.S = R.H.S Why did you plug n=1 for the last two, wouldn't it be easier just to plug n=0 for all and see that A has no solution? Just want to know if there was any specific reason why you did so Thank you Cheers J PS. Would be nice if we could get this question in code format! Kudos [?]: 770 [0], given: 355 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7863 Kudos [?]: 18465 [0], given: 237 Location: Pune, India Re: Exponents [#permalink] ### Show Tags 15 Jan 2014, 19:55 jlgdr wrote: nverma wrote: marcusaurelius wrote: Each of the following equations has at least one solution EXCEPT –2^n = (–2)^-n 2^-n = (–2)^n 2^n = (–2)^-n (–2)^n = –2^n (–2)^-n = –2^-n IMHO A a) –2^n = (–2)^-n –2^n = 1/(–2)^n –2^n * (–2)^n = 1, Keep it. Let's solve the other options..!! b) 2^-n = (–2)^n 1/2^n = (–2)^n 1 = (–2)^n * (2^n) For n=0, L.H.S = R.H.S c) 2^n = (–2)^-n 2^n = 1/ (–2)^n (2^n) * (–2)^n = 1 For n=0, L.H.S = R.H.S d) (–2)^n = –2^n (–2)^n / –2^n = 1 For n=1, L.H.S = R.H.S e) (–2)^-n = –2^-n 1/ (–2)^n = 1/–2^n For n=1, L.H.S = R.H.S Why did you plug n=1 for the last two, wouldn't it be easier just to plug n=0 for all and see that A has no solution? Just want to know if there was any specific reason why you did so Thank you Cheers J PS. Would be nice if we could get this question in code format! We need to find the equation that has no solution. What we are trying to do is find at least one solution for 4 equations. The fifth one will obviously not have any solution and will be our answer. Options (D) and (E) do not have 0 as a solution. So you try n = 1 on (A), (D) and (E). n = 1 is still not a solution for (A) but it is for (D) and (E). (D) (–2)^n = –2^n When you put n = 0, you get (-2)^0 = -2^0 1 = -1 which doesn't hold. So you try n = 1 (–2)^1 = -2^1 -2 = -2 n = 1 is a solution. Same logic for (E) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: No solution N: Manhattan GMAT test [#permalink]

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24 Aug 2014, 15:52
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Re: Each of the following equations has at least one solution [#permalink]

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08 Sep 2015, 07:44
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Re: Each of the following equations has at least one solution [#permalink]

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11 Sep 2016, 06:38
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Re: Each of the following equations has at least one solution [#permalink]

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31 Aug 2017, 16:58
Quote:
I mean that −xy−xy always means −(xy)−(xy). If it's supposed to mean (−x)y(−x)y, then it would be represented this way.

This means that any time we see a negative sign, it is isolated. So -2^2 in gemat does not equal 4 but -4?

I'm confused now as in all questions when you see -2^2, you will consider it 4 not -4

Please clarify what you think Brent.

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Re: Each of the following equations has at least one solution   [#permalink] 31 Aug 2017, 16:58

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