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Each of the following equations has at least one solution EX

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Senior Manager
Joined: 05 Oct 2008
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Each of the following equations has at least one solution EX [#permalink]

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17 Oct 2009, 07:06
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Question Stats:

40% (01:13) correct 60% (01:43) wrong based on 60 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: each-of-the-following-equations-has-at-least-one-solution-94119.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Nov 2013, 06:32, edited 3 times in total.
Renamed the topic, edited the question and added the OA.

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Senior Manager
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17 Oct 2009, 07:13
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Hey study.... are some of these options supposed to be exponents as the title suggests?
I think I've seen this before and a lot of the questions were written in exponent form.

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17 Oct 2009, 07:37
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yes...thanks for prompting me.

The exponent signs got missed out in the copy paste. Now the question reads well.

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17 Oct 2009, 18:38
1
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Plug in n=0

First are fine,
Last two 1 = -1 and 1 = -1 require closer look

Plug in n=1 in last two
d) (-2)^1 = -1 * 2^1 wrong
e) (–2)^-1 = –2^-1 -1/2 = -1/2 correct

Answer is D becase n=0 is sol-on for A,B,C and n=1 is solution for E
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17 Oct 2009, 23:45
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Hi study,
I guess something is wrong in options D and E.Both look the same to me..Pls put parenthesis for D and E so that it is a bit more clear

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18 Oct 2009, 03:33
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Hi Economist,

There is nothing wrong with E.

You are correct - It is the same and therefore it will have at least one answer to satisfy the equation. Hence E cannot be the correct answer choice. Hope this helps..

Anyone able to solve this??

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18 Oct 2009, 17:48
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study 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

A. –2^n = (–2)^-n
–2^n = 1/(–2)^n
(–2^n) (–2)^n = 1
None of the value of n, the equation is never valid..

B. 2^-n = (–2)^n
1/2^n = (–2)^n
1 = 2^n (–2)^n
If n is 0, then the equation is solved.

C. 2^n = (–2)^-n
2^n = 1/(–2)^n
2^n (–2)^n = 1
Same as B: If n is 0, then the equation is solved.

D. (–2)^n = –2^n
(–2)^n + 2^n = 0
n can have any non-even integer value.
If n = 1, the equation is valid.
If n = 3, the equation is valid and so on....

E. (–2)^-n = –2^-n
1/(–2)^n = 1/–2^n
–2^n = (–2)^n
n can have any non-even integer value.
If n = 1, the equation is valid.
If n = 3, the equation is valid and so on....

Got A but was bit confused..
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18 Oct 2009, 18:25
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GMAT TIGER wrote:

A. –2^n = (–2)^-n
–2^n = 1/(–2)^n
(–2^n) (–2)^n = 1
None of the value of n, the equation is never valid..

Got A but was bit confused..

I think the equation valid for n=0? Am I missing something

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18 Oct 2009, 19:24
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hgp2k wrote:
GMAT TIGER wrote:

A. –2^n = (–2)^-n
–2^n = 1/(–2)^n
(–2^n) (–2)^n = 1
None of the value of n, the equation is never valid..

Got A but was bit confused..

I think the equation valid for n=0? Am I missing something

A. –2^n = (–2)^-n
–2^n = 1/(–2)^n
(–2^n) (–2)^n = 1
(–2^0) (–2)^0 = 1
–1 (1) = 1
–1 = 1 .................not valid....
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18 Oct 2009, 21:28
1
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hgp2k wrote:
GMAT TIGER wrote:

A. –2^n = (–2)^-n
–2^n = 1/(–2)^n
(–2^n) (–2)^n = 1
None of the value of n, the equation is never valid..

Got A but was bit confused..

I think the equation valid for n=0? Am I missing something

thats what I was telling about parenthesis, may be because the format is not proper we are not able to understand.

Here, LHS is -ve of 2^n..so if n=0, we have -1 on LHS
RHS is (-2) raised to (-n).

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23 Oct 2009, 23:10
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It would have been better if the formula code were used!

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Current Student
Joined: 06 Sep 2013
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14 Nov 2013, 05:57
study 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

I tried the same method of plugging 0 and then trying to decide between D and E but not quite sure if its the correct way
Any experts want to give this one a shot?

Cheers
J

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Math Expert
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Posts: 42607

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14 Nov 2013, 06:32
jlgdr wrote:
study 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

I tried the same method of plugging 0 and then trying to decide between D and E but not quite sure if its the correct way
Any experts want to give this one a shot?

Cheers
J

OPEN DISCUSSION OF THIS QUESTION IS HERE: each-of-the-following-equations-has-at-least-one-solution-94119.html
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Re: Exponents   [#permalink] 14 Nov 2013, 06:32
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