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# Each of the following equations has at least one solution EXCEPT

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

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12 May 2010, 09:12
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Difficulty:

95% (hard)

Question Stats:

32% (02:02) correct 68% (02:05) wrong based on 512 sessions

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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. $$(–2)^{-n} = –2^{-n}$$
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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12 May 2010, 14:01
1
4
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 $$(-2)^{-n}=(-2)^{0}=1$$.
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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12 May 2010, 10:53
4
2
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
##### General Discussion
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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12 May 2010, 11:09
all seems to have n=0 as solution....? whats the OA
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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12 May 2010, 15:13
1
yes right i didnt read it closely
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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14 May 2010, 06:56
2
Hi,

it's really elegant question

0 is a solution for second and third equations.
1 is a solution for the last two equations.

So answer is A. Really if n is not equal to 0 then absolute value of left part is greater than 1 and right part is less than 1. In case when n is equal to 0 we will get -1=1.
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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14 May 2010, 12:49
great explanation Bunuel, thanks
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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15 Jun 2010, 13:15
Thanks for merging.

Do you mean a negative number raised to the power of 0 yields -1?? I didn't know that!
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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15 Jun 2010, 13:32
3
study wrote:
Thanks for merging.

Do you mean a negative number raised to the power of 0 yields -1?? I didn't know that!

No.

Any number to the power of zero equals to 1 (except 0^0: 0^0 is undefined for GMAT and not tested).

The point here is that $$-2^n$$ means $$-(2^n)$$ and not $$(-2)^n$$. So for $$n=0$$ --> $$-2^n=-(2^n)=-(2^0)=-(1)$$. But if it were $$(-2)^n$$, then for $$n=0$$ --> $$(-2)^0=1$$.

Hope it's clear.
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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15 Jun 2010, 22:56
So how do you know that the point here is that -2^n means -(2^n) and not (-2)^n

The actual question has no parenthesis. This is tricky!
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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16 Jun 2010, 05:36
study wrote:
So how do you know that the point here is that -2^n means -(2^n) and not (-2)^n

The actual question has no parenthesis. This is tricky!

I mean that $$-x^y$$ always means $$-(x^y)$$. If it's supposed to mean $$(-x)^y$$, then it would be represented this way.
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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06 Dec 2010, 08:21
2
SubratGmat2011 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

Can somebody plz help me out what is the approch for this type of problems?

The first and most straight forward approach that comes to mind is that I can see most of these equations will have n = 0 or n = 1 as a solution.
Except for the very first one:
n = 0: -2^0 = -1 while (-2)^(-0) = 1
n = 1: -2^1 = -2 while (-2)^-n = -1/2

For all other options, n = 0 or 1 satisfies the equation.
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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19 Jan 2011, 22:03
1
1
Lets look at each choice -

A –2^n = (–2)^-n
=>(-1).(2)^n = 1/(-2)^n
=>(-1).(2)^n.(-2)^n = 1
=>(-1).(2)^n.(-1)^n.(2)^n = 1
=>(-1).(-1)^n.(2)^2n = 1
Above cannot be true for any value of n (No solution - answer)

B 2^-n = (–2)^n
=>1/(2)^n = (-2)^n
=>1=(-1)^n.(2)^n.(2)^n
=>1=(-1)^n.(2)^2n
Above is true for n=0, so it has atleast one solution

C 2^n = (–2)^-n
=>(2)^n = 1/(-2)^n
Rest of the steps Similar to option B

D (–2)^n = –2^n
=>(-1)^n. (2)^n = (-1).(2)^n
=>(-1)^n = (-1)
Above is true for all odd values of n

E (–2)^-n = –2^-n
=>1/[(-1)^n. (2)^n] = (-1)/(2)^n
=>1/[(-1)^n] = (-1)
=>1/(-1)^n = -1
Above is true for all odd values of n

I hope this helps.
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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18 Feb 2011, 10:22
i guess the only real way to solve it under 2 min is to plug in 0/1...

if u start with choosing 1 here as first step is not good... choosing 0 is canceling 3 choices quickly...
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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05 May 2011, 17:26
lets pick a value for n.

n = 0

A. cannot be true as we get -1 on LHS and 1 on RHS ( as anything to the power of 0 is 1)
B. true (LHS = RHS = 1)
C. true (LHS = RHS = 1)
D. true (LHS = RHS = 1)
E. true (LHS = RHS = 1)

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

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05 May 2011, 22:42
for B and C n = 0
for D and E n = 1.

A prevails.
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Re: Each of the following equations has at least one solution EXCEPT  [#permalink]

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18 Aug 2012, 15:14
1
arichardson26 wrote:
Each of the following 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

B, C have can be equated by using n=0
D and E have external/independent -ve signs, so 0 wont help, but using n= +1 for D and -1 for E will equate the sides.

Took more than 2 mins
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Re: Each of the following equations 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 EXCEPT  [#permalink]

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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?

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

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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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Re: Each of the following equations has at least one solution EXCEPT &nbs [#permalink] 14 Oct 2012, 22:12

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