Question

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Answer 1

$3^x\cdot 8^{\frac{x}{x+1}}=36\\ \\ \log_{9}(3^x\cdot 8^{\frac{x}{x+1}}) = \log_{9}36\\ \\ \log_{9}3^x +\log_{9}8^{\frac{x}{x+1}} = \log_{9}(9\cdot 4)\\\ \\ x\log_{9}3 + \dfrac{x}{x+1}\log_{9}8 = 1+\log_{9}4\Big|\cdot(x+1) \\ \\ \dfrac{x^2+x}{2}+x\log_{9}8 = x+1+(x+1)\log_{9}4 \\ \\ x^2+x+2x\log_{9}8=2x+2+2(x+1)\log_{9}4 \\ \\ x^2+x(1+2\log_{9}8-2-2\log_{9}4) - (2+2\log_{9}4) = 0\\ \\ x^2+(2\log_{9}2 - 1)x - (2\log_{9}4+2) = 0\\ \\x^2+(\log_{9}4 - 1)x - (2\log_{9}4+2) = 0\\ \\\\\text{Notez: }\,\,\log_{9}4 = a>0 \\\\ \\ x^2+(a-1)x-2(a+1) = 0\\ \\ \Delta = (a-1)^2+8(a+1) = a^2-2a+1+8a+8 = a^2+6a+9 = \\ = (a+3)^2$

$x_{1,2} = \dfrac{1-a\pm (a+3)}{2}\\ \\\\\Rightarrow x_1 = \dfrac{1-a-a-3}{2} = -a-1 = -\log_{9}4-1\\\\ \Rightarrow x_2 = \dfrac{1-a+a+3}{2} =2\\ \\ \\ \Rightarrow \boxed{S = \Big\{-\log_{9}4-1;\,2\Big\}}$

Answer 2

Explicație pas cu pas:

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