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)
