Cho các số thực dương a, b thỏa mãn log4a=log6b=log9(4a−5b)−1. Đặt T=b/a

Hướng dẫn giải:

Đáp án D.

Giả sử: \({{\log }_{4}}a={{\log }_{6}}b={{\log }_{9}}(4a-5b)-1=t\)\(\Rightarrow \left\{ \begin{align}  & a={{4}^{t}} \\  & b={{6}^{t}} \\  & 4a-5b={{9}^{t+1}} \\ \end{align} \right.\)

Khi đó:  \( {{4.4}^{t}}-{{5.6}^{t}}={{9.9}^{t}}\Leftrightarrow 4{{\left( \frac{4}{9} \right)}^{t}}-5{{\left( \frac{6}{9} \right)}^{t}}=9 \)

 \( \Leftrightarrow 4{{\left( \frac{2}{3} \right)}^{2t}}-5{{\left( \frac{2}{3} \right)}^{t}}-9=0\Leftrightarrow \left[ \begin{align}  & {{\left( \frac{2}{3} \right)}^{t}}=\frac{9}{4}={{\left( \frac{2}{3} \right)}^{-2}} \\  & {{\left( \frac{2}{3} \right)}^{t}}=-1\text{(lo }\!\!{}^\text{1}\!\!\text{ i)} \\ \end{align} \right. \)

 \( \Leftrightarrow t=-2\Rightarrow T=\frac{b}{a}={{\left( \frac{6}{4} \right)}^{t}}={{\left( \frac{3}{2} \right)}^{-2}}=\frac{4}{9}\in \left( 0;\frac{1}{2} \right) \).