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