No Solution N

[atoledo]

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

According to the CAT, the correct answer is 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. "

I'm not sure I understand why though. Wouldn't n = 0 make all these equations equal?

[jeromecukier]

-2^n should be interpreted as -(2^n) and not as (-2)^n.

in printed mathematics, the unary operator minus (-) has a lower precedence than exponentiation, so 2^n should be computed before negation.

so -2^0 = -1.
if they meant powers of a negative number they would have had to put that number between parentheses.

[RonPurewal]

-2^n should be interpreted as -(2^n) and not as (-2)^n.

in printed mathematics, the unary operator minus (-) has a lower precedence than exponentiation, so 2^n should be computed before negation.

so -2^0 = -1.
if they meant powers of a negative number they would have had to put that number between parentheses.

very nice explanation.