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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