selango wrote: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
(A) -2^n is always negative for all values of n, while (-2)^-n is positive for even values and negative for odd integers.
If n = 0, then -2^n = -1 and (-2)^0 = 1, so for n = 0, it does not hold true.
If n = 1, then -2^n = -2 and (-2)^-n = -1/2, so -2^n is not equal to (-2)^-n and if n = 2, then -2^n = -4 and (-2)^-n = 1/4, again-2^n is not equal to (-2)^-n. This equation gives no solution for any value of n.
(B) If n = 0, then 2^-n = 2^0 =1 and (-2)^n = (-2)^0 = 1. So for n = 0, it holds true.
(C) If n =0, 2^n == 2^0 = 1 and (-2)^-n = (-2)^0 = 1. So for n = 0, it holds true.
(D) If n = 0, (-2)^n = (-2)^0 = 1 and -2^n = -2^0 = -1. But for any odd value, say n = 1, (-2)^1 = -2 and -2^n = -2^1 = -2. So, (-2)^n = -2^n holds true for odd values.
(E) If n = 0, (-2)^-n = (-2)^0 = 1 and -2^-n = -2^0 = -1. If n = 1, (-2)^-1 = -1/2 and -2^-n = -2^-1 = -1/2. So, (-2)^n = -2^n holds true for odd values.
[spoiler]
The correct answer is (A).[/spoiler]
Rahul Lakhani
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