Question

Mn giải giúp mình câu này với

Lo​g₃(36-3^x+4)=1-x

Answer 1

ĐKXĐ: \(36-3^{x+4}>0\Rightarrow x< log_3\dfrac{4}{9}\)

\(log_3\left(36-3^{x+4}\right)=1-x\Leftrightarrow36-3^{x+4}=3^{1-x}\)

\(\Leftrightarrow36.3^x-81.3^{2x}-3=0\)

Đặt \(3^x=t>0\Rightarrow-81t^2+36t-3=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{1}{3}\\t=\dfrac{1}{9}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3^x=\dfrac{1}{3}\\3^x=\dfrac{1}{9}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)