Giải bài 2.15 trang 109 - SBT Giải tích lớp 12
Tính
\(a)\,\dfrac 1 2\log_7{36}-\log_7{14}-3\log_7{\sqrt[3]{21}}\)
\(b)\,\dfrac{\log_2{24}-\dfrac 1 2 \log_272}{\log_318-\dfrac1 3\log _372}\)
\(c)\,\dfrac{\log_24+\log_2\sqrt{10}}{\log_2{20}+3\log_22}\)
Hướng dẫn:
Xem lại các quy tắc tính logarit của tích, logarit của thương và các công thức biến đổi.
a)
\(\,\,\,\,\,\dfrac 1 2\log_7{36}-\log_7{14}-3\log_7{\sqrt[3]{21}}\\ =\log_7\sqrt{36}-\log_714-3.\dfrac 1 3 \log_721\\ =\log_76-\log_714-\log_721\\ =\log_7{\dfrac{6}{14.21}}=\log_7{\dfrac 1 {49}}=-2\)
b)
\(\begin{align} & \dfrac{{{\log }_{2}}24-{{\log }_{2}}\sqrt{72}}{{{\log }_{3}}18-{{\log }_{3}}\sqrt[3]{72}} \\ & =\dfrac{{{\log }_{2}}\dfrac{24}{\sqrt{72}}}{{{\log }_{3}}\dfrac{18}{\sqrt[3]{72}}} \\ & =\dfrac{{{\log }_{2}}\dfrac{24}{3\sqrt{8}}}{{{\log }_{3}}\dfrac{18}{2\sqrt[3]{9}}} \\ & =\dfrac{{{\log }_{2}}\dfrac{8}{\sqrt{8}}}{{{\log }_{3}}\dfrac{9}{\sqrt[3]{9}}} \\ & =\dfrac{{{\log }_{2}}{{2}^{3-\frac{3}{2}}}}{{{\log }_{3}}{{3}^{2-\frac{2}{3}}}}=\dfrac{\dfrac{3}{2}}{\dfrac{4}{3}}=\dfrac{9}{8} \\ \end{align}\)
c)
\(\dfrac{\log_24+\log_2\sqrt{10}}{\log_2{20}+3\log_22}\\ \begin{align} & =\frac{{{\log }_{2}}\left( 4.\sqrt{10} \right)}{{{\log }_{2}}\left( {{20.2}^{3}} \right)} \\ & =\frac{{{\log }_{2}}\left( {{2}^{2}}{{.2}^{\frac{1}{2}}}{{.5}^{\frac{1}{2}}} \right)}{{{\log }_{2}}\left( {{2}^{2}}{{.5.2}^{3}} \right)} \\ & =\frac{{{\log }_{2}}{{2}^{\frac{5}{2}}}+{{\log }_{2}}{{5}^{\frac{1}{2}}}}{{{\log }_{2}}{{2}^{5}}+{{\log }_{2}}5} \\ & =\frac{\dfrac{1}{2}\left( {{\log }_{2}}{{2}^{5}}+{{\log }_{2}}5 \right)}{{{\log }_{2}}{{2}^{5}}+{{\log }_{2}}5}=\frac{1}{2} \\ \end{align} \)