Question 982391
<pre>
Notice that I've changed the numbers, so you'll have to go
through it, using this as a model.  Yours is done EXACTLY 
the same way:

{{{-3 + 2*ln(x-3) = ln(x)}}}

{{{-3 + 2*ln(x-3) = ln(x)}}}

{{{2*ln(x-3)-ln(x) = 3}}}

{{{ln(x-3)^2-ln(x) = 3}}}

{{{ln((x-3)^2/x)=e}}}

{{{(x-3)^2/x=e^3}}}

{{{(x-3)^2=e^3*x}}}

{{{x^2-6x+9=e^3*x}}}

{{{x^2-6x+9-e^3*x=0}}}

{{{x^2-6x-e^3*x+9=0}}}

{{{x^2-x(6+e^3)+9=0}}}

{{{x^2-(6+e^3)x+9=0}}}

{{{a=1}}}, {{{b=-(6+e^3)}}}, {{{c=9)}}}

{{{x = (-b +- sqrt( b^2-4ac ))/(2a) }}}

{{{x = (-(-(6+e^3)^"") +- sqrt((-(6+e^3)^"")^2-4(1)(9) ))/(2(1)) }}}


{{{x = ((6+e^3) +- sqrt((6+e^3)^2-36 ))/2 }}}

{{{x = (6+e^3 +- sqrt((36+12e^3+e^6)-36 ))/2 }}}

{{{x = (6+e^3 +- sqrt(36+12e^3+e^6-36 ))/2 }}}

{{{x = (6+e^3 +- sqrt(12e^3+e^6 ))/2 }}}

{{{x = (6+e^3 +- sqrt(e^3(12+e^3)))/2}}}

Using the +, that's approximately 25.7358
Using the -, that's approximately 0.3497

However since there is ln(x-2) in the original equation, the
value of x-2 must be positive, or x > 2, so we may not use 
the minus sign.

Solution:
x = (6+e^3+sqrt(e^3(12+e^3)))}}}

Now use the above as a model and do your problem.

Edwin</pre>