Bài tự luận · Bài 5
Dạng 1. Thực hiện các phép tính lôgarit
Bài tự luận Lôgarit · Bài 5
(CTST11) Tính giá trị của các biểu thức sau:
a) $\log_{9}\frac{1}{27}$;
b) $\log_{8}9\cdot \log_{27}\frac{1}{16}$;
c) $\log_{4}27\cdot \log_{3}5\cdot \log_{25}8$.
a) $\log_{9}\frac{1}{27}$;
b) $\log_{8}9\cdot \log_{27}\frac{1}{16}$;
c) $\log_{4}27\cdot \log_{3}5\cdot \log_{25}8$.
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Lời giải
a) $\log_{9}\frac{1}{27}=\frac{\log_{3}\frac{1}{27}}{\log_{3}9}=\frac{\log_{3}3^{-3}}{\log_{3}3^{2}}=-\frac{3}{2}$;
b) $\begin{array}{ll} \log_{8}9\cdot \log_{27}\frac{1}{16} & =\frac{\log_{2}9}{\log_{2}8}\cdot \frac{\log_{2}\frac{1}{16}}{\log_{2}27}=\frac{\log_{2}3^{2}}{\log_{2}2^{3}}\cdot \frac{\log_{2}2^{-4}}{\log_{2}3^{3}} \\ & =\frac{2\log_{2}3}{3\log_{2}2}\cdot \frac{-4\log_{2}2}{3\log_{2}3}=\frac{2}{3}\cdot \frac{-4}{3}=-\frac{8}{9} \\ \end{array}$
c) $\begin{array}{ll} \log_{4}27\cdot \log_{3}5\cdot \log_{25}8 & =\frac{\log_{2}27}{\log_{2}4}\cdot \frac{\log_{2}5}{\log_{2}3}\cdot \frac{\log_{2}8}{\log_{2}25} \\ & =\frac{\log_{2}3^{3}}{\log_{2}2^{2}}\cdot \frac{\log_{2}5}{\log_{2}3}\cdot \frac{\log_{2}2^{3}}{\log_{2}5^{2}} \\ & =\frac{3\log_{2}3}{2\log_{2}2}\cdot \frac{\log_{2}5}{\log_{2}3}\cdot \frac{3\log_{2}2}{2\log_{2}5}=\frac{9}{4}. \\ \end{array}$
b) $\begin{array}{ll} \log_{8}9\cdot \log_{27}\frac{1}{16} & =\frac{\log_{2}9}{\log_{2}8}\cdot \frac{\log_{2}\frac{1}{16}}{\log_{2}27}=\frac{\log_{2}3^{2}}{\log_{2}2^{3}}\cdot \frac{\log_{2}2^{-4}}{\log_{2}3^{3}} \\ & =\frac{2\log_{2}3}{3\log_{2}2}\cdot \frac{-4\log_{2}2}{3\log_{2}3}=\frac{2}{3}\cdot \frac{-4}{3}=-\frac{8}{9} \\ \end{array}$
c) $\begin{array}{ll} \log_{4}27\cdot \log_{3}5\cdot \log_{25}8 & =\frac{\log_{2}27}{\log_{2}4}\cdot \frac{\log_{2}5}{\log_{2}3}\cdot \frac{\log_{2}8}{\log_{2}25} \\ & =\frac{\log_{2}3^{3}}{\log_{2}2^{2}}\cdot \frac{\log_{2}5}{\log_{2}3}\cdot \frac{\log_{2}2^{3}}{\log_{2}5^{2}} \\ & =\frac{3\log_{2}3}{2\log_{2}2}\cdot \frac{\log_{2}5}{\log_{2}3}\cdot \frac{3\log_{2}2}{2\log_{2}5}=\frac{9}{4}. \\ \end{array}$