Cho {log _3}15 = a tính A = {log _{25}}15 theo a.​

Ta có: \(a = {\log _3}15 \Rightarrow {\log _3}5 + {\log _3}3 = a \Rightarrow {\log _3}5 = a - 1\)

\({\log _{25}}15 = \frac{{{{\log }_3}15}}{{{{\log }_3}25}} = \frac{{{{\log }_3}\left( {3.5} \right)}}{{{{\log }_3}{5^2}}} = \frac{{1 + {{\log }_3}5}}{{2.{{\log }_3}5}} = \frac{{1 + a - 1}}{{2.\left( {a - 1} \right)}} = \frac{a}{{2.\left( {a - 1} \right)}}.\)