Lời giải:Ta có: Gọi \(H\left( {{x}_{0}};0 \right).\) Khi đó \(A\left( {{x}_{0}};{{\log }_{a}}{{x}_{0}} \right);B\left( {{x}_{0}};{{\log }_{b}}{{x}_{0}} \right)\)
\(AH=\left| {{\log }_{a}}{{x}_{0}} \right|;BH=\left| {{\log }_{b}}{{x}_{0}} \right|\)
Do \(3HA=4HB\Leftrightarrow 3\left| {{\log }_{a}}{{x}_{0}} \right|=4\left| {{\log }_{b}}{{x}_{0}} \right|\)
Dựa vào đồ thị ta thấy: \(3\left| {{\log }_{a}}{{x}_{0}} \right|=4\left| {{\log }_{b}}{{x}_{0}} \right|\Leftrightarrow 3{{\log }_{a}}{{x}_{0}}=-4{{\log }_{b}}{{x}_{0}}\)
Đặt \(3{{\log }_{a}}{{x}_{0}}=-4{{\log }_{b}}{{x}_{0}}=t.\) Ta có
\(3{\log _a}{x_0} = - 4{\log _b}{x_0} = t \Leftrightarrow \left\{ \begin{array}{l}
{\log _a}{x_0} = \frac{t}{3}\\
{\log _b}{x_0} = - \frac{t}{4}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{a^{\frac{t}{3}}} = {x_0}\\
{b^{ - \frac{t}{4}}} = {x_0}
\end{array} \right.\)
\(\Leftrightarrow {{a}^{\frac{t}{3}}}={{b}^{-\frac{t}{4}}}\Leftrightarrow {{a}^{\frac{t}{3}}}=\frac{1}{{{b}^{\frac{t}{4}}}}\Leftrightarrow {{a}^{\frac{t}{3}}}.{{b}^{\frac{t}{4}}}=1\Leftrightarrow {{a}^{4}}.{{b}^{3}}=1.\)
