Question

Salut. O ecuație logarimtica, dacă se poate.... ​

Answer 1

$\log_{2}^2 x+(x-1)\log_{2}x = 6-2x\\ \\ \text{Notez: }\log_{2}x = t \Rightarrow x = 2^t,\,\,\,x > 0\\ \\ t^2+(2^t-1)t=6-2\cdot 2^t\\ t^2+2^tt-t-6+2\cdot 2^t = 0\\ t^2+2^t(t+2) - t-6 = 0\\ t^2+2^t(t+2)-(t+2)-4 = 0\\ t^2-4+2^t(t+2)-(t+2) = 0\\ (t-2)(t+2)+2^t(t+2)-(t+2) = 0\\ (t+2)(t-2+2^t-1) = 0\\ (t+2)(2^t+t-3) = 0$

$\\\boxed{1}\,\,\,t+2 = 0 \Rightarrow t= -2 \Rightarrow\\ \Rightarrow \log_{2}x = -2 \Rightarrow x = 2^{-2} \Rightarrow \boxed{x = \dfrac{1}{4}}\\\\\boxed{2}\,\,\,2^t+t-3 = 0 \Rightarrow \\ \Rightarrow 2^t = 3-t \Rightarrow t = 1\Rightarrow \log_{2}x = 1 \Rightarrow \boxed{x = 2}$

$\\\Rightarrow \boxed{\boxed{S = \left\{\dfrac{1}{4};\, 2\right\}}}$