Question

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

Answer 1

$displaystyle 27^{frac{x-2}{3}}-9^{x-1}=2*3^{2x-1}-2*3^{3x-1}\\frac{3^{x}}{9}}-3^{2x-2}=frac{2*3^{2x}}{3}-frac{2*3^{3x}}{3}\\frac{3^{x}-3^{2x}}{9}=frac{2*3^{2x}-2*3^{3x}}{3}\\3^{x}-3^{2x}=6*3^{2x}-6*3^{3x}\\3^{x}-3^{x})^{2}=6*((3^{x})^{2}-(3^{x})^{3})\\3^{x}(1-3^{x})=6*3^{2x}(1-3^{x})\\3^{x}=y\\y-y^{2}=6y^{2}(1-y)\y-y^{2}-6y^{2}+6y^{3}=0\y(6y^{2}-7y+1)=0\\yneq0\\6y^{2}-7y+1=0\D=b^{2}-4ac=49-24=25\\y_{1,2}=frac{-bбsqrt{D}}{2a}$

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

Ответ: {0; -log₃6}

Answer 2

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