2 б)log x+ 3log, x+2=0 2 2

б) $$log_{\frac{1}{2}}^2{x} + 3log_{\frac{1}{2}}{x} + 2 = 0$$

Пусть $$t = log_{\frac{1}{2}}{x}$$, тогда:

$$t^2 + 3t + 2 = 0$$

$$D = 3^2 - 4 \cdot 1 \cdot 2 = 9 - 8 = 1$$

$$t_1 = \frac{-3 + \sqrt{1}}{2} = \frac{-3 + 1}{2} = -1$$

$$t_2 = \frac{-3 - \sqrt{1}}{2} = \frac{-3 - 1}{2} = -2$$

Обратная замена:

$$log_{\frac{1}{2}}{x} = -1$$ => $$x = (\frac{1}{2})^{-1} = 2$$

$$log_{\frac{1}{2}}{x} = -2$$ => $$x = (\frac{1}{2})^{-2} = 4$$

Ответ: x = 2, x = 4