Question

Cho x,y là số thực dương \(log_9x=log_6y=log_4\left(\dfrac{x+y}{6}\right)\).Tính \(\dfrac{x}{y}\)

Answer 1

Lời giải:

Đặt \(\log_9x=\log_6y=\log_4\left(\frac{x+y}{6}\right)=t\)

\(\Rightarrow \left\{\begin{matrix} x=9^t\\ y=6^t\\ x+y=6.4^t\end{matrix}\right.\)

\(\Rightarrow 9^t+6^t=6.4^t\)

\(\Leftrightarrow \left(\frac{9}{6}\right)^t+1=6.\left(\frac{4}{6}\right)^t\)

\(\Leftrightarrow \left(\frac{3}{2}\right)^t+1=6.\left(\frac{2}{3}\right)^t\)

Đặt \(\left(\frac{3}{2}\right)^t=a\Rightarrow a+1=6.\frac{1}{a}\)

\(\Leftrightarrow a^2+a-6=0\Leftrightarrow a=2\) hoặc $a=-3$

Mà \(a>0\Rightarrow a=2\)

Ta có: \(\frac{x}{y}=\frac{9^t}{6^t}=\left(\frac{9}{6}\right)^t=\left(\frac{3}{2}\right)^t=a=2\)

Vậy \(\frac{x}{y}=2\)