Biễu diễn log _2}3 theo a= {log _12}8

Ta có:  \({\log _{12}}18 = \frac{{{{\log }_2}18}}{{{{\log }_2}12}} = \frac{{{{\log }_2}\left( {{{2.3}^2}} \right)}}{{{{\log }_2}({2^2}.3)}} = \frac{{1 + 2{{\log }_2}3}}{{2 + {{\log }_2}3}}\)

Đặt \(t = {\log _2}3\), ta có:

\({\log _{12}}18 = a = \frac{{1 + 2x}}{{2 + x}}\)

\(\Rightarrow a(2 + x) = 1 + 2x \Rightarrow x(a - 2) = 1 - 2a\)

\(\Rightarrow {\log _2}3 = x = \frac{{1 - 2a}}{{a - 2}}\)