[tex]\ln(5x-1)=\ln(3x+2)[/tex]
First note that the left side exists only for [tex]5x-1>0[/tex], or [tex]x>\dfrac15[/tex], and the right side exists only for [tex]3x+2>0[/tex], or [tex]x>-\dfrac23[/tex]. Then any solution to this equation is extraneous if we find [tex]x\le\dfrac15[/tex].
Write both sides as powers of [tex]e[/tex] to eliminate the logarithms:
[tex]e^{\ln(5x-1)}=e^{\ln(3x+2)}\implies5x-1=3x+2[/tex]
Simplify and solve:
[tex]2x=3\implies x=\dfrac32[/tex]
No extraneous solutions here because the one found is larger than 1/5.
