#f(x)=frac{(e^(2x-3))(lnx)}{x^3}#
#f(x)=(e^(2x-3))(lnx)(x^(-3))#
To find the equation of the tangent line at x=2, we need to find #f(2)# and #f'(2)#. Use the product rule to find #f'(x):#
#f'(x)=(e^(2x-3))(lnx)(-3x^(-4))+(e^(2x-3))(1/x)(x^(-3))+(2e^(2x-3))(lnx)(x^(-3))#
#f'(x)=frac{-3e^(2x-3)(lnx)}{x^4}+frac{e^(2x-3)}{x^4}+frac{2e^(2x-3)(lnx)color(red)(*x)}{x^3color (red)(*x)}#
#f'(x)=frac{-3e^(2x-3)(lnx)+e^(2x-3)+2xe^(2x-3)(lnx)}{x^4}#
#f'(x)=frac{(e^(2x-3))(-3lnx+1+2xlnx)}{x^4}#
#f'(2)=frac{e^(2color(red)((2))-3)(-3lncolor(red)((2))+1+2color(red)((2))lncolor(red)((2)))}{2^4}#
#f'(2)=frac{e(1+ln2)}{16}#
#f'(2)=frac{e+eln2}{16}#
f(2)#=frac{(e^(2(2)-3))(ln(2))}{(2)^3}#
#f(2)=frac{eln2}{8}#
Equation of tangent line (point-slope form):
#y-frac{eln2}{8}=frac{e+eln2}{16}(x-2)#
#y=(frac{e+eln2}{16})x-2(frac{e+eln2}{16})+frac{eln2}{8}#
#y=(frac{e+eln2}{16})x+(frac{eln2color(red)(*2)}{8color(red)(*2)}+frac{-2e-2eln2}{16})#
#y=(frac{e+eln2}{16})x+(frac{-2e}{16})#
#y=(frac{e+eln2}{16})x-e/8#
