Find the sum of all values of x such that $2^{x^2-3x-2} = 4^{x - 4} \cdot 8^x$.
+2407 2^(x^2 - 3x - 2) = 2^(2*(x-4)) * 2^(3x)
2^(x^2 - 3x - 2) = 2^(5x - 8)
x^2 - 3x - 2 = 5x - 8
x^2 - 8x + 6 = 0
Using vieta's, -b/a = --8/1 = 8.
=^._.^=
+37159 2^(x2-3x-2) = (2^2)x-4 * 2^3x equate the exponents
x^2 -3x -2 = 2x-8+3x
x^2 -8x +6 Use quadratic formula to find x = 4 +- sqrt(10) summed = 8
