Each of the following equations has at least one solution EXCEPT	Pick:
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
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 answer is –2^n = (–2)^-n
<result>A</result>
Each of the following equations has at least one solution EXCEPT