Đáp án:
Giải thích các bước giải:
$\eqalign{ & a = {\log _2}6 = {\log _2}(2.3) = 1 + {\log _2}3 \cr & \Rightarrow a - 1 = {\log _2}3 \cr & \Rightarrow \frac{{a - 1}}{a} = \frac{{{{\log }_2}3}}{{{{\log }_2}6}} = {\log _6}3 \cr & b = {\log _2}7 \cr & \Rightarrow \frac{{a - 1}}{b} = \frac{{{{\log }_2}3}}{{{{\log }_2}7}} = {\log _7}3 \cr} $
Khi đó:
$\eqalign{ & {\log _{18}}42 = {\log _{18}}(6.7) = {\log _{18}}6 + {\log _{18}}7 \cr & = \frac{1}{{{{\log }_6}18}} + \frac{1}{{{{\log }_7}18}} = \frac{1}{{{{\log }_6}({3^2}.2)}} + \frac{1}{{{{\log }_7}({3^2}.2)}} \cr & = \frac{1}{{{{\log }_6}{3^2} + {{\log }_6}2}} + \frac{1}{{{{\log }_7}{3^2} + {{\log }_7}2}} \cr & = \frac{1}{{2{{\log }_6}3 + {{\log }_6}2}} + \frac{1}{{2{{\log }_7}3 + {{\log }_7}2}} \cr & = \frac{1}{{2{{\log }_6}3 + \frac{1}{{{{\log }_2}6}}}} + \frac{1}{{2{{\log }_7}3 + \frac{1}{{{{\log }_2}7}}}} \cr} $
$ = \frac{1}{{2.\frac{{a - 1}}{a} + \frac{1}{a}}} + \frac{1}{{2.\frac{{a - 1}}{b} + \frac{1}{b}}} = \frac{a}{{2a - 1}} + \frac{b}{{2a - 1}} = \frac{{a + b}}{{2a - 1}}$
