\(\ln(x^2+x+5)=\ln(-3x+1)\)
First note that for both sides to be defined, we require \(x^2+x+5>0\) and \(-3x+1>0\). The first condition is met for all \(x\), and the other requires that \(x<\dfrac13\).
Now,
\(e^{\ln(x^2+x+5)}=e^{\ln(-3x+1)}\)
\(x^2+x+5=-3x+1\)
\(x^2+4x+4=0\)
\((x+2)^2=0\)
\(x=-2\)
Since \(-2<\dfrac13\), this is the only solution.
