### Instruction: Find the sum of all values of $x$ such that $2^{x^2-3x-2} = 4^{x - 4}$.### Response: I notice that both sides of the equation have powers of 2, so I can rewrite them using the same base: $2^{x^2-3x-2} = (2^2)^{x - 4}$.
Using the exponent rule $(a^b)^c = a^{bc}$, I can simplify the right side: $2^{x^2-3x-2} = 2^{2(x - 4)}$.
Now I have the same base on both sides, so I can equate the exponents and get a quadratic equation: $x^2 - 3x - 2 = 2x - 8$.
Moving everything to one side, I get $x^2 - 5x + 6 = 0$.
I can factor this into $(x - 2)(x - 3) = 0$.
Using the zero product property, I get two possible solutions for $x$: $x = 2$ or $x = 3$.
I can check that both solutions work by plugging them back into the original equation and verifying that they are true.