Tính giá trị của các biểu thức sau: log3 9/10 + log3 30

Giải SBT Toán 11 Bài 2: Phép tính lôgarit

Bài 3 trang 13 SBT Toán 11 Tập 2: Tính giá trị của các biểu thức sau:

a) ${\log }_3\frac{9}{10}+{\log }_{3}30$;

b) log₅ 75 + log₅ 3;

c) ${\log }_3\frac{5}{9}-2{\log }_3\sqrt{5}$;

d) 4log₁₂ 2 + 2log₁₂ 3;

e) $2{\log }_{5}2-{\log }_{5}4\sqrt{10}+{\log }_5\sqrt{2}$;

g) ${\log }_3\sqrt{3}-{\log }_3\sqrt[3]{9}+2{\log }_3\sqrt[4]{27}$.

Lời giải:

a) ${\log }_3\frac{9}{10}+{\log }_{3}30={\log }_3\left(\frac{9}{10}.30\right)={\log }_3{3}^3=3$;

b) ${\log }_{5}75-{\log }_{5}3={\log }_5\frac{75}{3}={\log }_{5}25={\log }_5{5}^2=2$;

c) ${\log }_3\frac{5}{9}-2{\log }_3\sqrt{5}={\log }_3\frac{5}{9}-{\log }_3{\left(\sqrt{5}\right)}^2$

= ${\log }_3\frac{5}{9}-{\log }_{3}5={\log }_3\left(\frac{5}{9}:5\right)={\log }_3{3}^{-2}=-2$;

d) $4{\log }_{12}2+2{\log }_{12}3={\log }_{12}{2}^4+{\log }_{12}{3}^2$

= ${\log }_{12}\left(2^4{.3}^3\right)={\log }_{12}{\left(4.3\right)}^2={\log }_{12}{12}^2=2$;

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) ${\log }_3\sqrt{3}-{\log }_3\sqrt[3]{9}+2{\log }_3\sqrt[4]{27}$

= ${\log }_3{3}^{\frac{1}{2}}-{\log }_3{3}^{\frac{2}{3}}+2{\log }_3{3}^{\frac{3}{4}}$

= $\frac{1}{2}-\frac{2}{3}+2.\frac{3}{4}=\frac{4}{3}$.