To find the equation of a tangent line, we will need the slope m and a point (x_{o}, y_{o}). Then, we will use any method to find an equation of a line (i.e. pointslope form, yintercept form). Here, we will have to find the slope and
then the point (e^{3},f(e3)).
Finding the slope: m = f'(e^{3})
f(x) = ln(x)
f(x) = ln(x) + ln(1) (property of logs)
f'(x) = 1/x
f'(e^{3}) = 1/e^{3}
Finding the point (e^{3},f(e^{3}))
f(x) = ln(x)
f(e^{3}) = ln((e^{3}))
f(e^{3}) = 3 (property of logs)
Our point is (e^{3}, 3)
Find the equation of a line
y  y_{o} = m(x  x_{o})
y = mx  mx_{o} + y_{o}
y = x/e^{3}  (1/e^{3})*(e^{3}) + 3
y = x/e^{3}  1 + 3
y = x/e^{3}  2 <==== equation of a tangent line
May 11

Xavier H.