Question 1109882
The equation has a few restrictions, but we can worry about it later,
as we may have to get rid of some extraneous solutions, anyway.
For now, we will assume {{{x<>0}}} , {{{x+4<>0}}} , etc.
{{{(ln(1+4/x)) ^2 +(ln(1-4/(x+4)))^2}}} {{{"="}}} {{{2(ln((3-x)/(x-1)))^2}}}
{{{(ln((x+4)/x)) ^2 +(ln((x+4-4)/(x+4)))^2 }}} {{{"="}}} {{{2(ln((3-x)/(x-1)))^2}}}
{{{(ln((x+4)/x)) ^2 +(ln(x/(x+4)))^2 }}} {{{"="}}} {{{2(ln((3-x)/(x-1)))^2}}}
{{{(ln((x+4)/x)) ^2 +(ln(x/(x+4)))^2+2*ln((x+4)/x)ln(x/(x+4))}}} {{{"="}}} {{{2(ln((3-x)/(x-1)))^2}}}
{{{(ln((x+4)/x) +ln(x/(x+4)))^2}}} {{{"="}}} {{{2(ln((3-x)/(x-1)))^2+2*ln((x+4)/x)ln(x/(x+4))}}}
{{{(ln((x+4)/x)*(x/(x+4)))^2}}} {{{"="}}} {{{2(ln((3-x)/(x-1)))^2+2*ln((x+4)/x)ln(x/(x+4))}}}
{{{(ln(1))^2}}} {{{"="}}} {{{2(ln((3-x)/(x-1)))^2+2*ln((x+4)/x)ln(x/(x+4))}}}
{{{0}}} {{{"="}}} {{{2(ln((3-x)/(x-1)))^2+2*ln((x+4)/x)ln(x/(x+4))}}}
{{{(ln((3-x)/(x-1)))^2+ln((x+4)/x)ln(x/(x+4))}}} {{{"="}}} {{{0}}}
{{{(ln((3-x)/(x-1)))^2)}}} {{{"="}}} {{{-ln((x+4)/x)ln(x/(x+4))}}}
{{{(ln((3-x)/(x-1)))^2)}}} {{{"="}}} {{{ln(x/(x+4))ln(x/(x+4))}}}
{{{(ln((3-x)/(x-1)))^2)}}} {{{"="}}} {{{(ln(x/(x+4)))^2}}}
If both logarithms have the same sign (both positive or both negative,
{{{ln((3-x)/(x-1)))}}} {{{"="}}} {{{ln(x/(x+4))}}} <--> {{{(3-x)/(x-1)}}} {{{"="}}} {{{x/(x+4)}}} .
Otherwise,
{{{ln((3-x)/(x-1)))}}} {{{"="}}} {{{-ln(x/(x+4))}}} <--> {{{ln((3-x)/(x-1)))}}} {{{"="}}} {{{-ln((x+4)/x)}}} <--> {{{(3-x)/(x-1)}}} {{{"="}}} {{{(x+4)/x}}} .
 
And we trudge on.
 
{{{(3-x)/(x-1)}}} {{{"="}}} {{{x/(x+4)}}}
{{{(3-x)(x+4)}}} {{{"="}}} {{{x(x-1)}}}
{{{12-x-x^2=x^2-x}}}
{{{12-x^2=x^2}}}
{{{12=2x^2}}}
{{{x^2=6}}}
{{{x=" " +- sqrt(6)}}}
For {{{highlight(x=sqrt(6))}}} ,
{{{3-x}}} , {{{x-1}}} , {{{x}}} , {{{x+4}}} , {{{(3-x)/(x-1)}}} , and {{{x/(x+4)}}} are all positive,
so it looks like a valid solution, and it verifies.
For {{{x=-sqrt(6)}}} , {{{x/(x+4)<0}}} ,
so {{{ln(1-4/(x+4))=ln(x/(x+4))}}} does not exist/is undefined,
so we know {{{x=-sqrt(6)}}} is not a solution.
I do not have to try to verify that.
 
{{{(3-x)/(x-1)}}} {{{"="}}} {{{(x+4)/x}}}
{{{(3-x)x}}} {{{"="}}} {{{(x+4)(x-1)}}}
{{{3x-x^2=x^2+3x-4}}}
{{{-x^2=x^2-4}}}
{{{2x^2=4}}}
{{{x^2=2}}}
{{{x=" " +- sqrt(2)}}}
For {{{highlight(x=sqrt(2))}}} ,
{{{3-x}}} , {{{x-1}}} , {{{x}}} , {{{x+4}}} , {{{(3-x)/(x-1)}}} , and {{{x/(x+4)}}} are all positive,
so it looks like a valid solution, and it verifies.
For {{{x=-sqrt(2)}}} , {{{x/(x+4)<0}}} ,
so {{{ln(1-4/(x+4))=ln(x/(x+4))}}} does not exist/is undefined,
so we know {{{x=-sqrt(2)}}} is not a solution.
I do not have to try to verify that.
 
HOW WE VERIFY:
For {{{highlight(x=sqrt(2))}}}
{{{(3-x)/(x-1)}}} {{{"="}}} {{{(3-sqrt(2))/(sqrt(2)-1)}}} {{{"="}}} {{{(3-sqrt(2))(sqrt(2)+1)/((sqrt(2)-1)(sqrt(2)+1))}}} {{{"="}}} {{{(3sqrt(2)+3-2-sqrt(2))/(2-1))}}} {{{"="}}} {{{2sqrt(2)+1}}}
 
{{{(x+4)/x}}} {{{"="}}} {{{(sqrt(2)+4)/sqrt(2)}}} {{{"="}}} {{{(sqrt(2)+4)sqrt(2)/(sqrt(2)sqrt(2))}}} {{{"="}}} {{{(2+4sqrt(2))/2}}} {{{"="}}} {{{2sqrt(2)+1}}}