Giải bài 2.17 trang 109 - SBT Giải tích lớp 12
a) Cho \( a = \log_315, b=\log_310\). Hãy tính \(\log_{\sqrt 3}50\), theo a và b
b) Cho \(a =\log_2 3, b=\log_3 5, c=\log_7 2\). Hãy tính \(\log_{140}63\) theo \(a, b, c.\)
a)
\(\begin{align} & {{\log }_{\sqrt{3}}}50=2.{{\log }_{3}}50=2.\left( {{\log }_{3}}5+{{\log }_{3}}10 \right) \\ & =2.\left( {{\log }_{3}}\dfrac{15}{3}+{{\log }_{3}}10 \right) \\ & =2.\left( {{\log }_{3}}15+{{\log }_{3}}10-{{\log }_{3}}3 \right) \\ & =2\left( a+b-1 \right) \\ \end{align}\)
b)
\(\begin{align} & {{\log }_{140}}63=\dfrac{{{\log }_{2}}63}{{{\log }_{2}}140} \\ & =\dfrac{{{\log }_{2}}\left( 9.7 \right)}{{{\log }_{2}}\left( {{2}^{2}}.5.7 \right)}=\dfrac{2{{\log }_{2}}3+{{\log }_{2}}7}{{{\log }_{2}}{{2}^{2}}+{{\log }_{2}}5+{{\log }_{2}}7} \\ & =\dfrac{2{{\log }_{2}}3+\dfrac{1}{{{\log }_{7}}2}}{2+\dfrac{{{\log }_{3}}5}{{{\log }_{3}}2}+\dfrac{1}{{{\log }_{7}}2}} \\ & =\dfrac{2a+\dfrac{1}{c}}{2+ab+\dfrac{1}{c}}=\dfrac{\dfrac{2ac+1}{c}}{\dfrac{2c+abc+1}{c}}=\dfrac{2{{a}}c+1}{abc+2c+1} \\ \end{align}\)