We will be using the following:
The quadratic formula:
ax^2 + bx + c = 0 => x = (-b +-sqrt(b^2-4ac))/(2a)
ln(a^b) = bln(a)
2^x = 2^(-x) - 3/2
Multiplying both sides by 2^x gives
(2^x)^2 = 1 - (3/2)2^x
=> (2^x)^2 + (3/2)2^x - 1 = 0
This is now a quadratic equation. Applying the quadratic formula, we obtain
2^x = (-3/2 +- sqrt(9/4 + 4))/2=(-3/2 +- 5/2)/2 = (-3+-5)/4
We know, however, that for all real-valued x, 2^x > 0 and so we can discard 2^x = -2 leaving us with 2^x = 1/2
Now, taking the natural log of both sides and applying the property of logarithms stated above,
ln(2^x) = ln(1/2)
=> xln(2) = ln(1/2)
=> x = ln(1/2)/ln(2)
For more complicated values, we would be done here, however as 1/2 = 2^(-1)
x = ln(2^(-1))/(ln(2)) = (-1ln(2))/(ln(2)) = -1
Thus x = -1
(Note that observing earlier that 1/2 = 2^-1 would have allowed us to say that x = -1 without any work with logarithms, however the above is a more generally applicable technique which works even when the values do not work out so nicely)
