2log(основание X)27-3log(основание 27)x=1

$2log_{x}2 - 3log_{27}x = 1 \\ \\ \dfrac{2}{log_{27}x} - 3log_{27}x = 1$
Пусть $t = log_{27}x$
$\dfrac{2}{t} - 3t = 1 \\ \\ 3t + 1 - \dfrac{2}{t} = 0 \\ \\ \dfrac{3t^2 + t - 2 }{t} = 0 \\ \\ 3t^2 + t - 2 = 0 \\ \\ D = 1 + 4 \cdot 2 \cdot 3 = 1 + 24 = 25 = 5^2 \\ \\ t_1 = \dfrac{-1 + 5 }{6} = \dfrac{2}{3} \\ \\ t_2 = \dfrac{-1 - 5}{6} = -1$
Обратная замена:
$log_{27}x = \dfrac{2}{3} \\ \\ 27^ \frac{2}{3}} = x \\ \\ 3^{2} = x \\ \\ \boxed{x = 9} \\ \\ log_{27}x = -1 \\ \\ \boxed{x = \dfrac{1}{27}}$