Cho ${\log _{12}}27 = a$. Tính \(T = {\log _{36}...

B

- Hướng dẫn giải

Ta có ${\log _{12}}27 = \frac{3}{{{{\log }_3}\left( {{2^2}.3} \right)}} = a$

Suy ra $\frac{3}{{1 + 2{{\log }_3}2}} = a$ hay ${\log _3}2 = \frac{{3 - a}}{{2a}}$ ( $a \ne 0$ vì $a = {\log _{12}}27 > {\log _{12}}1$ ).

Khi đó ${\log _{36}}24 = \frac{{{{\log }_3}24}}{{{{\log }_3}36}} = \frac{{3{{\log }_3}2 + 1}}{{2{{\log }_3}2 + 2}} = \frac{{1 + \frac{{9 - 3a}}{{2a}}}}{{2 + \frac{{6 - 2a}}{{2a}}}} = \frac{{9 - a}}{{6 + 2a}}$