Rút gọn biểu thức

Giải Toán 12 Bài 3: Lôgarit

Bài 3 trang 68 Toán lớp 12 Giải tích:

a) ${\log }_{3}6.{\log }_{8}9.{\log }_{6}2$;

b) ${\log }_a{b}^2+{\log }_{a^2}{b}^4$;

Lời giải:

a)

${\log }_{3}6.{\log }_{8}9.{\log }_{6}2=\left({\log }_{3}6.{\log }_{6}2\right).{\log }_{8}9={\log }_{3}2.{\log }_{2^3}{3}^2={\log }_{3}2.\left(\frac{1}{3}.2{\log }_{2}3\right)=\frac{2}{3}\left({\log }_{3}2.{\log }_{2}3\right)=\frac{2}{3}{\log }_{3}3=\frac{2}{3}$

b) 

${\log }_a{b}^2+{\log }_{a^2}{b}^4={\log }_a{b}^2+\frac{1}{2}{\log }_a{\left(b^2\right)}^2={\log }_a{b}^2+\frac{1}{2}.2{\log }_a{b}^2={\log }_a{b}^2+{\log }_a{b}^2=2{\log }_a{b}^2=4{\log }_a\left|b\right|$

Cách khác:

${\log }_a{b}^2+{\log }_{a^2}{b}^4=2{\log }_a\left|b\right|+4.\frac{1}{2}{\log }_a\left|b\right|=2{\log }_a\left|b\right|+2{\log }_a\left|b\right|=4{\log }_a\left|b\right|$