Question

Решить логарифмическое уравнение log3x+log9x+log27x=11/12

Answer 1

$\displaystyle log_{3}x+log_{9}x+log_{27}x= \frac{11}{12} \\ \\ log_{3}x+log_{3^{2}}x+log_{3^{3}}x= \frac{11}{12} \\ \\ log_{3}x+ \frac{1}{2}log_{3}x+ \frac{1}{3}log_{3}x= \frac{11}{12} \\ \\ \frac{6}{6}log_{3}x+ \frac{3}{6}log_{3}x+ \frac{2}{6}log_{3}x= \frac{11}{12} \\ \\ \frac{11}{6}log_{3}x= \frac{11}{12} \\ \\ log_{3}x= \frac{11}{12}* \frac{6}{11} \\ \\ log_{3}x= \frac{1}{2} \\ \\ x=3^{1/2} \\ \\ x= \sqrt{3}$

Ответ: {√3}