Решить показательные уравнения.

Решите задачу:

$9^{x}-2^{x+0.5} = 2^{x+3.5}-3^{2x-1}\\ 3^{2x}+3^{2x}/3 = 2^{x+3.5}+2^{x+0.5}\\ 9^x(1+1/3) = 2^x\sqrt{2}(2^3+1)\\ (9/2)^x=9\div4/3\\ (9/2)^x = 27/(2\sqrt{2}) = (9/2)^{3/2}\\ x=3/2\\\\ 4^x-3^{x-0.5}=3^{x+0.5}-2^{2x-1}\\ 4^x+4^x/2 =3^{x+0.5}+3^{x-0.5}\\ 4^x\cdot3/2 = 3^x(\sqrt{3}+1/\sqrt{3})\\ (4/3)^x =\frac{2}{3}(\sqrt{3}+1/\sqrt{3}) = \frac{2}{3}\frac{4}{\sqrt{3}} = (4/3)^{3/2}\\ x=3/2$