Bài tự luận · Bài 8
Dạng 1. Thực hiện các phép tính lôgarit
Bài tự luận Lôgarit · Bài 8
(CTST11) Tính giá trị của các biểu thức sau:
a) $\log_{3}\frac{9}{10}+\log_{3}30$;
b) $\log_{5}75-\log_{5}3$;
c) $\log_{3}\frac{5}{9}-2\log_{3}\sqrt{5}$;
d) $4\log_{12}2+2\log_{12}3$;
e) $2\log_{5}2-\log_{5}4\sqrt{10}+\log_{5}\sqrt{2}$;
g) $\log_{3}\sqrt{3}-\log_{3}\sqrt[3]{9}+2\log_{3}\sqrt[4]{27}$
a) $\log_{3}\frac{9}{10}+\log_{3}30$;
b) $\log_{5}75-\log_{5}3$;
c) $\log_{3}\frac{5}{9}-2\log_{3}\sqrt{5}$;
d) $4\log_{12}2+2\log_{12}3$;
e) $2\log_{5}2-\log_{5}4\sqrt{10}+\log_{5}\sqrt{2}$;
g) $\log_{3}\sqrt{3}-\log_{3}\sqrt[3]{9}+2\log_{3}\sqrt[4]{27}$
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Lời giải
a) $\log_{3}\frac{9}{10}+\log_{3}30=\log_{3}\left( \frac{9}{10}\cdot 30 \right)=\log_{3}3^{3}=3$;
b) $\log_{5}75-\log_{5}3=\log_{5}\frac{75}{3}=\log_{5}25=\log_{5}5^{2}=2$;
c) $\log_{3}\frac{5}{9}-2\log_{3}\sqrt{5}=\log_{3}\frac{5}{9}-\log_{3}(\sqrt{5})^{2}=\log_{3}\frac{5}{9}-\log_{3}5=\log_{3}\left( \frac{5}{9}:5 \right)=\log_{3}\frac{1}{9}$ $=\log_{3}3^{-2}=-2$;
d) $4\log_{12}2+2\log_{12}3=\log_{12}2^{4}+\log_{12}3^{2}=\log_{12}\left( 2^{4}\cdot 3^{2} \right)=\log_{12}(4\cdot 3)^{2}$ $=\log_{12}12^{2}=2$
e) $2\log_{5}2-\log_{5}4\sqrt{10}+\log_{5}\sqrt{2}=\log_{5}2^{2}-\log_{5}4\sqrt{10}+\log_{5}\sqrt{2}$
$=\log_{5}4-\log_{5}4\sqrt{10}+\log_{5}\sqrt{2}=\log_{5}\frac{4\sqrt{2}}{4\sqrt{10}}=\log_{5}\frac{1}{\sqrt{5}}=\log_{5}5^{-\frac{1}{2}}=-\frac{1}{2}$;
g) $\log_{3}\sqrt{3}-\log_{3}\sqrt[3]{9}+2\log_{3}\sqrt[4]{27}=\log_{3}3^{\frac{1}{2}}-\log_{3}3^{\frac{2}{3}}+2\log_{3}3^{\frac{3}{4}}$
$=\frac{1}{2}-\frac{2}{3}+2\cdot \frac{3}{4}=\frac{4}{3}$.
b) $\log_{5}75-\log_{5}3=\log_{5}\frac{75}{3}=\log_{5}25=\log_{5}5^{2}=2$;
c) $\log_{3}\frac{5}{9}-2\log_{3}\sqrt{5}=\log_{3}\frac{5}{9}-\log_{3}(\sqrt{5})^{2}=\log_{3}\frac{5}{9}-\log_{3}5=\log_{3}\left( \frac{5}{9}:5 \right)=\log_{3}\frac{1}{9}$ $=\log_{3}3^{-2}=-2$;
d) $4\log_{12}2+2\log_{12}3=\log_{12}2^{4}+\log_{12}3^{2}=\log_{12}\left( 2^{4}\cdot 3^{2} \right)=\log_{12}(4\cdot 3)^{2}$ $=\log_{12}12^{2}=2$
e) $2\log_{5}2-\log_{5}4\sqrt{10}+\log_{5}\sqrt{2}=\log_{5}2^{2}-\log_{5}4\sqrt{10}+\log_{5}\sqrt{2}$
$=\log_{5}4-\log_{5}4\sqrt{10}+\log_{5}\sqrt{2}=\log_{5}\frac{4\sqrt{2}}{4\sqrt{10}}=\log_{5}\frac{1}{\sqrt{5}}=\log_{5}5^{-\frac{1}{2}}=-\frac{1}{2}$;
g) $\log_{3}\sqrt{3}-\log_{3}\sqrt[3]{9}+2\log_{3}\sqrt[4]{27}=\log_{3}3^{\frac{1}{2}}-\log_{3}3^{\frac{2}{3}}+2\log_{3}3^{\frac{3}{4}}$
$=\frac{1}{2}-\frac{2}{3}+2\cdot \frac{3}{4}=\frac{4}{3}$.