3*4^x-3^x+1/2=2^2x Решите уравнение

Решите задачу:

$3*4^{(x)}-3^{(x+\frac{1}{2})}=2^{(2x)} \\ 3*4^{(x)}-3^{(x+\frac{1}{2})}=4^{(x)} \\ \\ 3*4^{(x)}-4^{(x)}=3^{(x+\frac{1}{2})} \\ (3-1)*4^{(x)}=3^{(x)}*3^{(\frac{1}{2})} \\ 2*4^{(x)}=3^{(x)}* \sqrt{3}(|:3^{(x)}and|: 2) \\ \frac{4^{(x)}}{3^{(x)}} = \frac{\sqrt{3}}{2} \\ (\frac{4}{3})^{(x)} = (\frac{2}{\sqrt{3}})^{(-1)} \\ (\frac{2}{\sqrt{3}})^{(2x)} = (\frac{2}{\sqrt{3}})^{(-1)} \\ \\ 2x=-1 \\ x=- \frac{1}{2}$