The value of $$x$$ satisfying $$4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$$ is of the form $$\frac{a}{b}$$ where $$\gcd(a, b) = 1$$. Then the value of $$\left(\frac{a+b}{a-b}\right)$$ is equal to

Transposingto get the same bases on either side of the equation we get,

$$2^{2x} + 2^{2x - 1} = 3^{x + 1/2} + 3^{x - 1/2}$$, that is $$\dfrac{3}{2} \cdot 2^{2x} = \dfrac{4}{\sqrt{3}} \cdot 3^{x}$$. 

This simplifies to 

$$2^{2x - 3} = 3^{\frac{2x-3}{2}}$$

$$\left(\dfrac{2}{\sqrt{3}}\right)^{2x - 3} = 1 \implies 2x - 3 = 0$$

$$x = \dfrac{3}{2}$$. 

This gives $$a = 3$$ and $$b = 2$$

$$\dfrac{a+b}{a-b} = \dfrac{5}{1} = 5$$.