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Each of the following equations has at least one solution

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Re: Each of the following equations has at least one solution [#permalink] New post 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: Exponents [#permalink] New post 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.


Thank you for your response.

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: Exponents [#permalink] New post 14 Oct 2012, 22:12
Expert's post
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 has one solution except? [#permalink] New post 17 Jan 2013, 06:27
I lost myself trying to solve this one with algebra; just plug in numbers in similar spots and you'll be fine.
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Re: Each of the following equations has at least one solution [#permalink] New post 26 Oct 2013, 03:40
Can somebody please explain if (–2)^n = 1 or -1 (if n=0)
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Re: Exponents [#permalink] New post 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!
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Re: Exponents [#permalink] New post 15 Jan 2014, 19:55
Expert's post
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)
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Re: Exponents   [#permalink] 15 Jan 2014, 19:55
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