Bài tự luận · Bài 10
Dạng 1. Thực hiện các phép tính lôgarit
Bài tự luận Lôgarit · Bài 10
(CTST11) Tính:
a) $\log_{3}5\cdot \log_{5}7\cdot \log_{7}9$;
b) $\log_{2}\frac{1}{25}\cdot \log_{3}\frac{1}{32}\cdot \log_{5}\frac{1}{27}$.
a) $\log_{3}5\cdot \log_{5}7\cdot \log_{7}9$;
b) $\log_{2}\frac{1}{25}\cdot \log_{3}\frac{1}{32}\cdot \log_{5}\frac{1}{27}$.
Xem lời giải
Lời giải
a) $\log_{3}5\cdot \log_{5}7\cdot \log_{7}9=\log_{3}5\cdot \frac{\log_{3}7}{\log_{3}5}\cdot \frac{\log_{3}9}{\log_{3}7}=\log_{3}3^{2}=2$;
b)
$\log_{2}\frac{1}{25}\cdot \log_{3}\frac{1}{32}\cdot \log_{5}\frac{1}{27}=\log_{2}5^{-2}\cdot \log_{3}2^{-5}\cdot \log_{5}3^{-3}$
$=(-2)\log_{2}5\cdot (-5)\log_{3}2\cdot (-3)\log_{5}3$
$=-30\log_{2}5\cdot \log_{3}2\cdot \log_{5}3$
$=-30\log_{2}5\cdot \frac{\log_{2}2}{\log_{2}3}\cdot \frac{\log_{2}3}{\log_{2}5}=-30$
b)
$\log_{2}\frac{1}{25}\cdot \log_{3}\frac{1}{32}\cdot \log_{5}\frac{1}{27}=\log_{2}5^{-2}\cdot \log_{3}2^{-5}\cdot \log_{5}3^{-3}$
$=(-2)\log_{2}5\cdot (-5)\log_{3}2\cdot (-3)\log_{5}3$
$=-30\log_{2}5\cdot \log_{3}2\cdot \log_{5}3$
$=-30\log_{2}5\cdot \frac{\log_{2}2}{\log_{2}3}\cdot \frac{\log_{2}3}{\log_{2}5}=-30$