SubratGmat2011 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

Can somebody plz help me out what is the approch for this type of problems?

Is the answer E ?
(-2)^-n +2^-n = 0
1/ (-2)^n + 1/ 2^n = 0 which can never be the case ......

As an approach , you have to evaluate every option till you hit the right one....

Hi all!

IMO A

let us plug in numbers 0 and 1

(E) (-2)^-n = -2^-n
n = 0
(-2)^-0 = -2^-0
1 = -1
n = 1
(-2)^-1 = -2^-1
-0.5 = -0.5
n = 1 is a solution, so this answer is incorrect
(D) (-2)^n = -2^n
n = 0
(-2)^0 = -2^0
1 = -1
n = 1
(-2)^1 = -2^1
-2 = -2
n = 1 is a solution, so this answer is incorrect
(C) 2^n = (-2)^-n
n = 0
2^0 = (-2)^-0
1 = 1
n = 0 is a solution, so this answer is incorrect
(B) 2^-n = (-2)^n
n = 0
2^-0 = (-2)^0
1 = 1
n = 0 is a solution, so this answer is incorrect
(A) -2^n = (-2)^-n
n = 0
-2^0 = (-2)^-0
-1 = 1
n = 1
-2^1 = (-2)^-1
-2 = -0.5
All other equations have at least one solution. This one does not have.
It is correct answer

Hope it helps!
Best,
Maciek

Thanks a Lot Maciek!!! It helps.