Question

27^(x-2/3)-9^(x-1)=2*3^(2x-1)-2*3^(3x-1)

Answer 1

$\displaystyle 27^{\frac{x-2}{3}}-9^{x-1}=2*3^{2x-1}-2*3^{3x-1}\\\\\frac{3^{x}}{9}}-3^{2x-2}=\frac{2*3^{2x}}{3}-\frac{2*3^{3x}}{3}\\\\\frac{3^{x}-3^{2x}}{9}=\frac{2*3^{2x}-2*3^{3x}}{3}\\\\3^{x}-3^{2x}=6*3^{2x}-6*3^{3x}\\\\3^{x}-3^{x})^{2}=6*((3^{x})^{2}-(3^{x})^{3})\\\\3^{x}(1-3^{x})=6*3^{2x}(1-3^{x})\\\\3^{x}=y\\\\y-y^{2}=6y^{2}(1-y)\\y-y^{2}-6y^{2}+6y^{3}=0\\y(6y^{2}-7y+1)=0\\\\y\neq0\\\\6y^{2}-7y+1=0\\D=b^{2}-4ac=49-24=25\\\\y_{1,2}=\frac{-bб\sqrt{D}}{2a}$

$\displaystyle y_{1}=1 \\\\ y_{2}= \frac{1}{6} \\ \\3^{x}=1 \\x_{1} =0 \\ \\3^{x}= \frac{1}{6} \\ \\x_{2}=-log_{3}6$

ответ: {0; -log₃6}