Question 1004909
Logarithm of a  quotient is the difference of the logarithms, so
{{{log(3,2x-3)-log(3,x+2)=red(3)}}}--->{{{log(3,(2x-3)/(x+2))=red(3)}}}
Logarithm of something is the exponent you have to apply the base to get that something, so if there is a solution,
{{{log(3,(2x-3)/(x+2))=red(3)}}}--->{{{(2x-3)/(x+2)=3^red(3)}}}--->{{{(2x-3)/(x+2)=27}}} .
The rest is simple algebra:
{{{(2x-3)/(x+2)=27}}}--->{{{(2x-3)=27(x+2)}}}--->{{{2x-3=27x+54}}}--->{{{2x-27x=54+3}}}--->{{{-25x=57}}}--->{{{x=57/(-25)}}}--->{{{x=-57/25=-
2.28}}}
The problem is that with {{{x=-2.28}}} , {{{system(2x-3=-7.56,x+2=-0.28)}}} ,
and logarithm of a negative number is undefined.
So, there are {{{highlight(no)}}}{{{highlight(solutions)}}} .
We could have started with "there are no solutions with
{{{system(2x-3<=0,"or",x+2<=0)}}}<--->{{{system(x<=3/2,"or",x<=-3)}}}<--->{{{x<=3/2}}} ".
For {{{x>3/2}}} both logarithms exist, and {{{log(3,2x-3)-log(3,x+2)=log(3,(2x-3)/(x+2))}}} .
Then, if you were in calculus class (and even if you weren't),
you would realize that in the domain of that function
{{{(2x-3)/(x+2)}}} increases with {{{x}}} , and
{{{log(3,(2x-3)/(x+2))}}} increases with {{{(2x-3)/(x+2)}}} , so
{{{log(3,(2x-3)/(x+2))}}} increases with {{{x}}} , and
{{{(2x-3)/(x+2)=(2x+4-4+3)/(x+2)=(2x+4)/(x+2)-1/(x+2)=2-1/(x+2)<2}}} ,
so {{{log(3,(2x-3)/(x+2))<log(3,2)<3}}} .
Graphing {{{log(3,(2x-3)/(x+2))}}} and {{{log(3,2)}}} we get 
{{{graph(300,300,-10,90,-5,5,log(3,2x-3)-log(3,x+2),log(3,2))}}}