What is the sum of all possible values of \(x\) for which \((-3)^{x^2 -2x-3}=27^{x+7}?\)

-3 raised to an odd power is always negative and raised to an even power, positive. And -3 raised to an even power is the same as 3 raised to that power, eg (-3)^2=9= 3^2.

But the right side is always positive, so X^2-2x-3 has to be even.

Since 27 is 3^3, the right side can be rewritten as 3^(3X+21). And the left side can be rewritten as 3^(X^2-2X-3).

So 3^(X^2-2X-3)= 3^(3X+21)

Equating the exponents:

X^2-2X-3= 3X+21

X^2-5X-24=0
Factoring:
(X-8)(X+3)=0

So, X=8 or -3

But need to check whether these meet the even criteria:
8^2-(2*8)-3 = 64-19= 45, an odd power, so 8 doesn't work.

Checking -3:

-3^2+6-3= 15-3=12, an even power, so -3 works.

Since only one solution, the sum of possible solutions is -3, A