I'm solving the following exponential equation
$$4^{x}-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2x-1}$$
(My attempt is below)
\begin{align}4^{x}-3^{x-\frac{1}{2}}&=3^{x+\frac{1}{2}}-2^{2x-1}\\\\ 4^{x}+2^{2x-1}&=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}}\\\\ 2^{2x}+2^{2x}\cdot2^{-1}&=3^{x}\cdot3^{\frac{1}{2}}+3^{x}\cdot3^{-\frac{1}{2}}\\\\ 2^{2x}(1+\frac{1}{2})&=3^{x}(\sqrt{3}+\frac{1}{\sqrt{3}})\\\\ \frac{3}{2}\cdot2^{2x}&=\frac{4\sqrt{3}}{3}\cdot3^{x}\\\\ 3\cdot2^{2x-1}&=4\sqrt{3}\cdot3^{x-1}\\\\ 2^{2x-3}&=3^{x-\frac{3}{2}}\end{align}
It is clear that $$x=\frac{3}{2}$$ is a solution because all exponential functions are equal to 1 when the exponent is 0. But how do I prove it's the only solution? My apologies if it's trivial.
Exponential functions are strictly increasing. At $x<3/2$ we have that $3^{x-\frac32} < 2^{2x-3}$ and at $x>3/2$ we have that $3^{x-\frac32} > 2^{2x-3}$. Using these observations We can say that $3/2$ is the only solution.