Question

\(log_{x^2}16+log_{2x}64=3\)

Answer 1

ĐKXĐ: \(x>0\) ; \(x\ne1\)

\(\Leftrightarrow\dfrac{1}{2}log_x2^4+log_{2x}2^6=3\)

\(\Leftrightarrow2log_x2+6log_{2x}2=3\)

\(\Rightarrow\dfrac{2}{log_2x}+\dfrac{6}{log_22x}=3\)

\(\Leftrightarrow\dfrac{2}{log_2x}+\dfrac{6}{log_2x+1}=3\)

Đặt \(log_2x=t\)

\(\Rightarrow\dfrac{2}{t}+\dfrac{6}{t+1}=3\)

\(\Rightarrow\left[{}\begin{matrix}t=2\\t=-\dfrac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}log_2x=2\\log_2x=-\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{\sqrt[3]{2}}\end{matrix}\right.\)