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:
$2^{2x}+2^{2x-1}=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}}$
$2^{2x}(1+2^{-1})=3^x\left(3^{\frac{1}{2}}+3^{-\frac{1}{2}}\right)$
$2^{2x}*\frac{3}{2}=3^x\left(\sqrt{3}+\frac{1}{\sqrt{3}}\right)$
$2^{2x}*\frac{3}{2}=3^x(\frac{\sqrt{3}^2+1}{\sqrt{3}})$
$2^{2x}*\frac{3}{2}=3^x*\frac{4\sqrt{3}}{3}$--> here I have rationalized the fraction
$\frac{2^{2x}}{3^x}=\frac{4\sqrt{3}}{3}*\frac{2}{3}$
$\left(\frac{2^2}{3}\right)^x=\frac{8\sqrt{3}}{9}$
Now I hope I didn't make an error. How should I proceed from this point? Is there another way around this problem?
I think you did not make a mistake.
Your last equation it's $$\left(\frac{2}{\sqrt3}\right)^{2x}=\left(\frac{2}{\sqrt3}\right)^3$$ because $$\frac{8\sqrt3}{9}=\frac{8\sqrt3}{\sqrt81}=\frac{8}{\sqrt{27}}=\frac{2^3}{\sqrt{3^3}}=\frac{2^3}{\left(\sqrt3\right)^3}=\left(\frac{2}{\sqrt3}\right)^3.$$