Problem Solving

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

plug in -1,1,0 (-ve, +ve odd and even) into answer choices and check
a) -2^n = (-2)^-n has no solution -> -2^-1=(-2)^+1, -2^1=(-2)^-1 and -2^0=(-2)^0 {pemdas https://www.beatthegmat.com/mba/2011/02/ ... arithmetic , -1*2^0=(-2)^0 <> -1=1}
IOM a

hi,
follow the link...
https://www.beatthegmat.com/at-least-one ... 66137.html

Also logically
option A is the only one which has no solution

imagine can - 2^n = (-1/2)^n ???
it will never be equal to each other...
best case... try putting n=0
u will get -1 = 1
otherwise though with odd powers of n u can match the sign on either side of the equation... but u will never be able to match the value 2 to 1/2

hope it helps...

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

Find the four answer choices that each have a 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.

Thanks for d help Guys:)