PS

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

Strategy:
Eliminate the four answer choices that DO have at least one solution.
Since a� = 1 and a¹ = a, n=0 and n=1 are likely to work in at least some of the answer choices.
Test n=0 and n=1 in the five answer choices.

A. –2^n = (–2)^-n
n=0:
-(2^0) = (-2)^-0
-1 = 1. Doesn't work.

n=1:
-(2^1) = (-2)^-1
-2 = -1/2. Doesn't work.
Hold onto A.

B. 2^-n = (–2)^n
n=0:
2^-0 = (-2)^0
1=1.
n=0 is a solution. Eliminate B.

C. 2^n = (–2)^-n
n=0:
2^0 = (-2)^-0
1=1.
n=0 is a solution. Eliminate C.

D. (–2)^n = –2^n
n=1:
(-2)^1 = -(2^1)
-2 = -2.
n=1 is a solution. Eliminate D.

E. (–2)^-n = –2^-n
n=1:
(-2)^(-1) = -(2^-1)
-1/2 = - 1/2
n=1 is a solution. Eliminate E.

The correct answer is A.

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

Option A

-2^n = (-2)^-n
-2^n = 1/(-2)^n
(-2)^n x (-2^n) = 1
(-1)^n x (2^n) x (-1) x (2^n) = 1
(-1)^(n+1) x (2^2n) = 1
This is impossible (in the first sight) for any value of n other than zero as (2^2n) will become a bigger number for any other values of n bringing much higher values for expression on Left Hand Side (LHS)
and
for n =0

LHS = (-1)^(0+1) x (2^20) and RHS = 1
LHS = -1 and RHS =1
and -1 is not equal to +1, therefore,
[spoiler]So no Solution Correct Option A[/spoiler]