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) The left side is always negative, while the right side is positive for even values of n and negative for odd values of n. Therefore, the two sides of this equation are reciprocals when n is odd, and opposite reciprocals when n is even; the absolute values won’t be the same unless n = 0, but the signs won’t be the same unless n is odd. Therefore, the equation has no solution.
(B) The left side is always positive, while the right side is positive for even values of n and negative for odd values of n. Therefore, the two sides of the equation are reciprocals when n is even, and opposite reciprocals when n is odd. The only solution to the equation is n = 0, which produces 1 on both sides.
(C) The left side is always positive, while the right side is positive for even values of n and negative for odd values of n. Therefore, the two sides of the equation are reciprocals when n is even, and opposite reciprocals when n is odd. The only solution to the equation is n = 0, which produces 1 on both sides.
(D) The left side is positive for even values of n and negative for odd values of n, while the right side is always negative; the absolute values of the two sides are always the same (= 2n). Therefore, any odd value of n will solve this equation.
(E) The left side is positive for even values of n and negative for odd values of n, while the right side is always negative; the absolute values of the two sides are always the same (= 2-n). Therefore, any odd value of n will solve this equation.
The correct answer is A.
Why isn't n=0 a viable solution for answer choice (A)? If you plug in n=0, don't you get 1=1?