Bài tự luận · Bài 11
Dạng 1. Thực hiện các phép tính lôgarit
Bài tự luận Lôgarit · Bài 11
(CD11) Cho $\log_{a}b=2$. Tính:
a) $\log_{a}\left( a^{2}b^{3} \right)$;
b) $\log_{a}\frac{a\sqrt{a}}{b\sqrt[3]{b}}$;
c) $\log_{a}(2b)+\log_{a}\left( \frac{b^{2}}{2} \right)$
a) $\log_{a}\left( a^{2}b^{3} \right)$;
b) $\log_{a}\frac{a\sqrt{a}}{b\sqrt[3]{b}}$;
c) $\log_{a}(2b)+\log_{a}\left( \frac{b^{2}}{2} \right)$
Xem lời giải
Lời giải
a) $\log_{a}\left( a^{2}b^{3} \right)=\log_{a}a^{2}+\log_{a}b^{3}=2+3\log_{a}b=2+3.2=8$.
b) $\log_{a}\frac{a\sqrt{a}}{b\sqrt[3]{b}}=\log_{a}a^{\frac{3}{2}}-\log_{a}b^{\frac{4}{3}}=\frac{3}{2}-\frac{4}{3}\log_{a}b=\frac{3}{2}-\frac{4}{3}\cdot 2=-\frac{7}{6}$.
c) $\log_{a}(2b)+\log_{a}\left( \frac{b^{2}}{2} \right)=\log_{a}2+\log_{a}b+\log_{a}b^{2}-\log_{a}2=3\log_{a}b=3.2=6$.
b) $\log_{a}\frac{a\sqrt{a}}{b\sqrt[3]{b}}=\log_{a}a^{\frac{3}{2}}-\log_{a}b^{\frac{4}{3}}=\frac{3}{2}-\frac{4}{3}\log_{a}b=\frac{3}{2}-\frac{4}{3}\cdot 2=-\frac{7}{6}$.
c) $\log_{a}(2b)+\log_{a}\left( \frac{b^{2}}{2} \right)=\log_{a}2+\log_{a}b+\log_{a}b^{2}-\log_{a}2=3\log_{a}b=3.2=6$.