Question 735805
Use the rule {{{lna + ln b = ln(ab)}}}

so {{{ln(4x-5)+ln(x-3)=ln((4x-5)(x-3))}}}

so now you have 
{{{ln((4x-5)(x-3))=ln15}}}
so {{{(4x-5)(x-3)=15}}}
{{{4x^2-5x-12x+15=15}}}
{{{4x^2-17x=0}}}
{{{x(4x-17)=0}}}
{{{x=0}}}or {{{x=17/4}}}

but must omit x=0 since it has us take ln of a negative number when substituted into 4x-5 or x-3, so it is not in the domain. 

so answer is x=17/4