Đề bài

a) ജCho \({\log _2}3 = a.\) Tính \({\log _{18}}72\) theo \(a.\)

b)  Cho \(\log 2 = a.\) Tính \({\log _{20}}50\) theo \(a.\)

Lời giải chi tiết

a) Ta có:\({\log _{18}}72 = \frac{{{{\log }_2}72}}{{{{\log }_2}18}} = \frac{{{{\log }_2}({2^3}{{.3}^2})}}{{{{\log }_2}({{2.3}^2})}} = \frac{{{{\log }_2}{2^3} + {{\log }_2}{3^2}}}{{{{\log }_2}2 + {{\log }_2}{3^2}}} = \frac{{3 + 2{{\log }_2}3}}{{1 + 2{{\log }_2}3}} = \frac{{3 + 2a}}{{1 + 2a}}.\)b) Ta có:\({\log _{20}}50 = \frac{{\log 50}}{{\log 20}} = \frac{{\log \left( {{{10}^2}{{.2}^{ - 1}}} \right)}}{{\log \left( {2.10} \right)}} = \frac{{\log {{10}^2} + \log {2^{ - 1}}}}{{\log 2 + \log 10}} = \frac{{2 - \log 2}}{{\log 2 + 1}} = \frac{{2 - a}}{{a + 1}}.\)