Bài 2.15 trang 109 SBT giải tích 12

Đề bài

Tính:

a) \(\displaystyle\frac{1}{2}{\log _7}36 - {\log _7}14 - 3{\log _7}\sqrt[3]{{21}}\)

b) \(\displaystyle\frac{{{{\log }_2}24 - \frac{1}{2}{{\log }_2}72}}{{{{\log }_3}18 - \frac{1}{3}{{\log }_3}72}}\)

c) \(\displaystyle\frac{{{{\log }_2}4 + {{\log }_2}\sqrt {10} }}{{{{\log }_2}20 + 3{{\log }_2}2}}\)

Lời giải chi tiết

a) \(\displaystyle{\log _7}\sqrt {36}  - {\log _7}14 - {\log _7}21\)\(\displaystyle = {\log _7}6 - {\log _7}14 - {\log _7}21\) \(\displaystyle = {\log _7}\left( {\frac{6}{{14}}:21} \right)\)

\(\displaystyle = {\log _7}\frac{1}{{49}} = {\log _7}\left( {{7^{ - 2}}} \right) =  - 2\)

b) \(\displaystyle\frac{{{{\log }_2}24 - {{\log }_2}\sqrt {72} }}{{{{\log }_3}18 - {{\log }_3}\sqrt[3]{{72}}}}\)\(\displaystyle = \frac{{{{\log }_2}\frac{{24}}{{\sqrt {72} }}}}{{{{\log }_3}\frac{{18}}{{\sqrt[3]{{72}}}}}} = \frac{{{{\log }_2}\frac{{24}}{{6\sqrt 2 }}}}{{{{\log }_3}\frac{9}{{\sqrt[3]{9}}}}}\) \(\displaystyle = \frac{{{{\log }_2}\left( {2\sqrt 2 } \right)}}{{{{\log }_3}{{\left( {\sqrt[3]{9}} \right)}^2}}} = \frac{{{{\log }_2}{2^{\frac{3}{2}}}}}{{{{\log }_3}{3^{\frac{4}{3}}}}}\) \(\displaystyle = \frac{3}{2}:\frac{4}{3} = \frac{9}{8}\)

c) \(\displaystyle\frac{{{{\log }_2}4 + {{\log }_2}\sqrt {10} }}{{{{\log }_2}20 + 3{{\log }_2}2}}\)\(\displaystyle = \frac{{{{\log }_2}{2^2} + {{\log }_2}\left( {{2^{\frac{1}{2}}}{{.5}^{\frac{1}{2}}}} \right)}}{{{{\log }_2}\left( {{2^2}.5} \right) + 3}}\)\(\displaystyle = \frac{{2 + \frac{1}{2} + \frac{1}{2}{{\log }_2}5}}{{2 + 3 + {{\log }_2}5}}\) \(\displaystyle = \frac{{\frac{5}{2} + \frac{1}{2}{{\log }_2}5}}{{5 + {{\log }_2}5}} = \frac{1}{2}\)