#lnx=3+ln(x-3)#
Subtract #ln(x-3)# from each side.
#lnx-ln(x-3)=3#
Recall that #lna-lnb=ln(a/b)#.
∴ #ln(x/(x-3)) = 3#
Convert the logarithmic equation to an exponential equation.
#e^ln(x/(x-3)) = e^3#
Remember that #e^lnx =x#, so
#x/(x-3)=e^3#
#x=e^3(x-3) = xe^3-3e^3#
#xe^3-x=3e^3#
#x(e^3-1)=3e^3#
#x=(3e^3)/(e^3-1)#
Check:
# lnx=3+ln(x-3)#
If #x=(3e^3)/(e^3-1)#,
#ln((3e^3)/(e^3-1)) = 3+ln((3e^3)/(e^3-1)-3)#
#ln(3e^3)-ln(e^3-1)=3+ln((3e^3-3(e^3-1))/(e^3-1))#
#ln(3e^3)-ln(e^3-1)=3+ln((color(red)(cancel(color(black)(3e^3)))-color(red)(cancel(color(black)(3e^3)))+3)/(e^3-1))#
#ln(3e^3)-color(red)(cancel(color(black)(ln(e^3-1))))=3+ln3-color(red)(cancel(color(black)(ln(e^3-1))))#
#ln3+3=3+ln3#
#x=(3e^3)/(e^3-1)# is a solution.
