Giải câu 3 bài: Lôgarit.
Ta có:
a) $\log _{3}6.\log _{8}9.\log _{6}2$
= $\log _{8}9.\log _{3}6.\log _{6}2$
= $\log _{8}9.\log _{3}2$
= $\frac{\log _{3}2}{\log _{8}9}$
= $\frac{2\log _{3}2}{3\log _{3}2}=\frac{2}{3}$
Vậy $\log _{3}6.\log _{8}9.\log _{6}2=\frac{2}{3}$
b) $\log _{a}b^{2}+\log _{a^{2}}b^{4}$
= $2\log _{a}.\left |b \right |+2\log _{a}\left |b \right |$
= $4\log _{a}\left |b \right |$
Vậy $\log _{a}b^{2}+\log _{a^{2}}b^{4}=4\log _{a}\left |b \right |$