“the sum of all possible values of x” can be a distraction for those looking for multiple such values because there can be only one or even no such value.
\((−3)^{(x^2 − 2x − 3)} = 27^{(x + 7)}\)
\((−3)^{(x^2 − 2x − 3)} = 3^{3(x + 7)}\)
We can see that the powers MUST be even for the equation to hold true. Let’s find the values of x that give such result.
\((x^2 − 2x − 3) = 3(x + 7)\)
\(x^2 − 2x − 3 = 3x + 21\)
\(x^2 – 5x -24 = 0\)
\((x - 8)(x + 3) = 0\)
So, x can be either -3 or 8
Let’s solve the powers for these values and check whether the result is even. Only -3 gives an even power, so it is the answer.
Hence
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