Question

(log9 27+log27 9)*12 Решите пожалуйста

Answer 1

$(log_{9} \ 27+ log_{27} \ 9) \cdot 12 = (log_{3^2} \ 3^3+ log_{3^3} \ 3^2) \cdot 12 = \\ (\frac {3}{2} log_{3} \ 3+ \frac {2}{3} log_{3} \ 3) \cdot 12 = (\frac {3}{2}+ \frac {2}{3} ) \cdot 12 = \frac {3}{2} \cdot 12 + \frac {2}{3} \cdot 12 = 18+8 =26$