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.
How do you approach this question .
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- rockeyb
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Nope thats it . I have checked it this is the correct question . BTW in D and E LHS and RHS are not sameliferocks wrote:Hmm...now for both D and E, LHS=RHS
is there something else missing?
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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?
(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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- rockeyb
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Thanks for your reply . Your answer is correct but I do not understand your explanation . Can you please elaborate on your solution .
Thanks .
Thanks .
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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?
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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-2^n = (-2)^-nrockeyb wrote:How did you get this ?liferocks wrote:For option A we get,
2^2n=(-1)^(n-1)..
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
"If you don't know where you are going, any road will get you there."
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