Phương trình \(\log_3\left(4^x+15.2^x+27\right)-2\log_3\left(4.2^x-3\right)=0\) có nghiệm là
\(x=\log_34\).\(x=\log_43\).\(x=\log_23\).\(x=\log_32\).Hướng dẫn giải:\(\log_3\left(4^x+15.2^x+27\right)-2\log_3\left(4.2^x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}4.2^x-3>0\\\log_3\left(4^x+15.2^x+27\right)=\log_3\left(4.2^x-3\right)^2\end{matrix}\right.\)
Đặt \(t=2^x>0\) ta được
\(\left\{{}\begin{matrix}t>\dfrac{3}{4}\\t^2+15t+27=16t^2-24t+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t>\dfrac{3}{4}\\15t^2-39t-18=0\end{matrix}\right.\)
\(\Leftrightarrow t=3\Leftrightarrow2^x=3\)
\(\Leftrightarrow x=\log_23.\)
