Problem Solving

Hi,
Can someone confirms that 2^-0 is not 2/0 but rather 1 just as per 2^0?

Each of the following equations has at least one solution EXCEPT

-2^n = (-2)^-n
2^-n = (-2)^n
2^n = (-2)^-n
(-2)^n = -2^n
(-2)^-n = -2^-n

Answer is a.
Thx

Thank you.
What do you mean by LHS?

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

For many test-takers, the most practical approach will be to eliminate the four answer choices that DO have at least one solution.
Values likely to work in a majority of the answer choices are n=0 and n=1.
Plug n=0 and n=1 into the 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.