Struggling to finish an exponential equation

I have this exponential equation I tried solving but I'm stuck at the end.

The problem is this:

$${4}^{x}-{3}^{x-\frac{1}{2}}={3}^{x+\frac{1}{2}}-{2}^{2x-1}$$

And my approach is this:

${2}^{2x}+{2}^{2x-1}={3}^{x+\frac{1}{2}}+{3}^{x-\frac{1}{2}}$

${2}^{2x}(1+{2}^{-1})={3}^{x}({3}^{\frac{1}{2}}+{3}^{-\frac{1}{2}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}(\sqrt{3}+\frac{1}{\sqrt{3}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}(\frac{{\sqrt{3}}^{2}+1}{\sqrt{3}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}\ast \frac{4\sqrt{3}}{3}$--> here I have rationalized the fraction

$\frac{{2}^{2x}}{{3}^{x}}=\frac{4\sqrt{3}}{3}\ast \frac{2}{3}$

${\left(\frac{{2}^{2}}{3}\right)}^{x}=\frac{8\sqrt{3}}{9}$

Now I hope I didn't make an error. How should I proceed from this point? Is there another way around this problem?

I have this exponential equation I tried solving but I'm stuck at the end.

The problem is this:

$${4}^{x}-{3}^{x-\frac{1}{2}}={3}^{x+\frac{1}{2}}-{2}^{2x-1}$$

And my approach is this:

${2}^{2x}+{2}^{2x-1}={3}^{x+\frac{1}{2}}+{3}^{x-\frac{1}{2}}$

${2}^{2x}(1+{2}^{-1})={3}^{x}({3}^{\frac{1}{2}}+{3}^{-\frac{1}{2}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}(\sqrt{3}+\frac{1}{\sqrt{3}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}(\frac{{\sqrt{3}}^{2}+1}{\sqrt{3}})$

${2}^{2x}\ast \frac{3}{2}={3}^{x}\ast \frac{4\sqrt{3}}{3}$--> here I have rationalized the fraction

$\frac{{2}^{2x}}{{3}^{x}}=\frac{4\sqrt{3}}{3}\ast \frac{2}{3}$

${\left(\frac{{2}^{2}}{3}\right)}^{x}=\frac{8\sqrt{3}}{9}$

Now I hope I didn't make an error. How should I proceed from this point? Is there another way around this problem?