Question

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

Answer 1

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

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

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

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

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

$log_{3}x = 1$

$x = 3$