Tính: log3 5. log5 7 .log7 9

Giải SBT Toán 11 Bài 2: Phép tính lôgarit

Bài 5 trang 13 SBT Toán 11 Tập 2: Tính:

a) log₃ 5. log₅ 7 .log₇ 9;

b) ${\log }_2\frac{1}{25}.{\log }_3\frac{1}{32}.{\log }_5\frac{1}{27}$.

Lời giải:

a) log₃ 5. log₅ 7 .log₇ 9;

= ${\log }_{3}5.\frac{{\log }_{3}7}{{\log }_{3}5}.\frac{{\log }_{3}9}{{\log }_{3}7}$

= $lo{g}_3{3}^2=2$.

b) ${\log }_2\frac{1}{25}.{\log }_3\frac{1}{32}.{\log }_5\frac{1}{27}={\log }_2{5}^{-2}.{\log }_3{2}^{-5}.{\log }_5{3}^{-3}$

⇔ (–2) log₂ 5. (–5)log₃ 2. (–3) . (–3) log₅ 3

⇔ –30 log₂ 5 . log₂ 3 . log₅ 3 log₇ 21

⇔ $-30{\log }_{2}5.\frac{{\log }_{2}2}{{\log }_{2}3}.\frac{{\log }_{2}3}{{\log }_{2}5}=-30$.