THe equation of a tangent line to the graph of y=lnx at x=e^2

I get the derivative but then to get the equation I plus it in

y-lnx=1/e^2(x-e^2)

and that gives me
y=x/e^2-1+lnx

but my book says it does not have a lnx in the final answer help

Question

THe equation of a tangent line to the graph of y=lnx at x=e^2

I get the derivative but then to get the equation I plus it in

y-lnx=1/e^2(x-e^2)

and that gives me 
y=x/e^2-1+lnx

but my book says it does not have a lnx in the final answer help

campbell_st · Answer

oops... the point the tangent passes through is 

$$x = e^2$$ then 
$$y = \ln(e^2)$$

you need to simplify this.... the slope of the tangent is correct

anonymous · Answer

so does that mean I derive ln(e^2)?

campbell_st · Answer

it means the point you use for the tangent is

$$(e^2, \ln(e^2)).... or... (e^2, 2)$$