РЕШИТЕ УРАВНЕНИЕ:1)log3x+log9x+log27x=одиннадцать двенадцатых

$log_{3}x +$ $log_{9}x +$ $log_{27}x = \frac{11}{12}$

$log_{3}x +$ $log_{ 3^{2}}x +$ $log_{3^{3}}x = \frac{11}{12}$

$log_{3}x +$ $\frac{1}{2}log_{3}x +$ $\frac{1}{3}log_{3}x = \frac{11}{12}$

$\frac{1}{12}log_{3}x +$ $\frac{6}{12}$ $log_{3}x + \frac{4}{12}log_{3}x =$ $\frac{11}{12}$

$\frac{11}{12}log_{3}x =$ $\frac{11}{12}$

$log_{3}x = 1$

$x = 3$