Here
for option D LHS=RHS so we cannot solve it.

Ans option D

I have edited the question have a look .

Hmm...now for both D and E, LHS=RHS

is there something else missing?

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

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?

Thanks for your reply . Your answer is correct but I do not understand your explanation . Can you please elaborate on your solution .

Thanks .

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?

liferocks wrote:For option A we get,

2^2n=(-1)^(n-1)..

How did you get this ?

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