study 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 = (–2)^-n

–2^n = 1/(–2)^n

(–2^n) (–2)^n = 1

None of the value of n, the equation is never valid..

B. 2^-n = (–2)^n

1/2^n = (–2)^n

1 = 2^n (–2)^n

If n is 0, then the equation is solved.

C. 2^n = (–2)^-n

2^n = 1/(–2)^n

2^n (–2)^n = 1

Same as B: If n is 0, then the equation is solved.

D. (–2)^n = –2^n

(–2)^n + 2^n = 0

n can have any non-even integer value.

If n = 1, the equation is valid.

If n = 3, the equation is valid and so on....

E. (–2)^-n = –2^-n

1/(–2)^n = 1/–2^n

–2^n = (–2)^n

n can have any non-even integer value.

If n = 1, the equation is valid.

If n = 3, the equation is valid and so on....

Got A but was bit confused..

_________________

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