Решите уравнение: [latex] 9^{x} - 2^{x-0.5} = 2^{x+2.5} - 3^{2x-1} [/latex]

[latex] 9^{x} - 2^{x-0.5} = 2^{x+2.5} - 3^{2x-1} \\\ 9^{x} +3^{2x-1} = 2^{x-0.5} + 2^{x+2.5} \\\ 9^x+ \dfrac{3^{2x}}{3} = \dfrac{2^x}{2^{0.5}} +2^x\cdot 2^{2.5} \\\\ 9^x+ \dfrac{9^x}{3} = \dfrac{2^x}{2^{0.5}} +2^x\cdot 2^{2.5} \\\\ 9^x+ \dfrac{1}{3} \cdot 9^x= \dfrac{\sqrt{2}}{ 2 } \cdot 2^x+4 \sqrt{2}\cdot 2^x \\\ (1+ \dfrac{1}{3}) \cdot 9^x= (\dfrac{\sqrt{2}}{ 2 } +4 \sqrt{2})\cdot 2^x[/latex] [latex] \dfrac{4}{3} \cdot 9^x= \dfrac{9\sqrt{2}}{ 2 } \cdot 2^x \\\ \dfrac{9^x}{2^x} = \dfrac{9\sqrt{2}}{ 2 } : \dfrac{4}{3} \\\ \left( \dfrac{9}{2} \right)^x= \dfrac{27\sqrt{2}}{ 8 } \\\ \left( \dfrac{9}{2} \right)^x= \dfrac{27}{ 4 \sqrt{2} } \\\ x=\log_{4.5}\frac{27}{ 4 \sqrt{2} } =\log_{4.5}(\frac{9}{ 2 }\cdot \frac{9}{ 2 }\cdot \frac{1}{3 \sqrt{2} } )=2+\log_{4.5}\frac{1}{3 \sqrt{2} } =2-\log_{4.5}3 \sqrt{2} [/latex] Ответ: [latex]2-\log_{4.5}3 \sqrt{2} [/latex]