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How do you approach this question .

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by rockeyb » Tue May 18, 2010 10:52 pm
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


Please explain .

Source : MGMAT.
Last edited by rockeyb on Tue May 18, 2010 11:19 pm, edited 1 time in total.
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by liferocks » Tue May 18, 2010 11:03 pm
Here
for option D LHS=RHS so we cannot solve it.

Ans option D
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by rockeyb » Tue May 18, 2010 11:20 pm
I have edited the question have a look .
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by liferocks » Tue May 18, 2010 11:26 pm
Hmm...now for both D and E, LHS=RHS

is there something else missing?
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by rockeyb » Tue May 18, 2010 11:32 pm
liferocks wrote:Hmm...now for both D and E, LHS=RHS

is there something else missing?
Nope thats it . I have checked it this is the correct question . BTW in D and E LHS and RHS are not same ;)
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by liferocks » Wed May 19, 2010 12:53 am
Ok. Lets try to solve one by one


(E)(-2)^-n = -2^-n
or (-1)^-n*(2)^-n=(-1)*2^-n
or (-1)^-n=-1..
or (-1)^n=-1..this is true for all odd n

(D)(-2)^n = -2^n
or (-1)^n*2^n=(-1)*2^n
or (-1)^n=-1..this is true for all odd n

(C)2^n = (-2)^-n
or 2^n={(-1)^n}*1/(2^n)
or 2^2n=(-1)^n..this is possible only when n=0

(B)2^-n = (-2)^n
or 1/2^n={(-1)^n}*(2^n)
or {(-1)^n}*(2^2n)=1..this is possible when n=0

(A)-2^n = (-2)^-n
or -2^n={(-1)^n}*1/(2^n)
or 2^2n=(-1)^(n-1)...this do not have any solution


Ans should be A :) what is OA btw?
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by rockeyb » Wed May 19, 2010 1:11 am
Thanks for your reply . Your answer is correct but I do not understand your explanation . Can you please elaborate on your solution .

Thanks .
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by liferocks » Wed May 19, 2010 1:44 am
For option A we get,

2^2n=(-1)^(n-1)..now for any real value of n LHS cannot be equal to RHS..because for all real n RHS is -1..and for no real n
2^2n will be -1
This is why I concluded that this equation does not provide any real solution for n.

Does this help?
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by rockeyb » Wed May 19, 2010 3:03 am
liferocks wrote:For option A we get,

2^2n=(-1)^(n-1)..
How did you get this ?
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by liferocks » Wed May 19, 2010 8:53 pm
rockeyb wrote:
liferocks wrote:For option A we get,

2^2n=(-1)^(n-1)..
How did you get this ?
-2^n = (-2)^-n

or (-1)*2^n=(-1)^-n*(2)^-n
or 2^n/2^-n=(-1)^-n/(-1)
or 2^{n-(-n)}=(-1)^(-n+1)
or 2^2n=(-1)^(1-n)..small typo in the bold part
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