The given equation is a quadratic equation in terms of (2^x)^2, so let's make a substitution to simplify it:
Let y = 2^x.
Substituting this into the equation, we get:
(y^2) - (7y) + 6 = 0.
Now, we can solve this quadratic equation for y by factoring or using the quadratic formula:
(y - 6)(y - 1) = 0.
From this, we find two possible values for y:
y = 6, y = 1.
Recalling the substitution, y = 2^x:
2^x = 6, 2^x = 1.
For the equation 2^x = 6, we take the logarithm of both sides to solve for x:
x = log2(6).
For the equation 2^x = 1, we know that any power of 2 equals 1 only when x = 0.
Thus, we have two possible solutions: x = log2(6) and x = 0.
To find the sum of the roots, we add these solutions:
Sum of the roots = log2(6) + 0.