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 $\log _{a}b^{2}+\log _{a^{2}}b^{4}=4\log _{a}\left |b  \right |$