ln(5x-2) = ln6 - ln(x-3)
Thanks!
Rewrite with you log rules to get
$\displaystyle \ln(5x-2)=\ln\left( \frac{6}{x-3}\right)$
Since both sides are the same natural log the arguments must be equal
$\displaystyle 5x-2=\frac{6}{x-3}$ Clearing the fraction we get
$\displaystyle (x-3)(5x-2)=6 \iff 5x^2-17x+6=6 \iff 5x^2-17x=0$
$\displaystyle x(5x-17)=0$ so we get the two solutions $\displaystyle x=0 \\\ x=\frac{17}{5}$
x cannot be zero why? hmmm but the other will work.
I hope this helps.
Good luck.