Can u find the equation of the tangent line to y=e^(3)+lnx when x=1?

Answer 1

y'(1) = slope of tangent line
y' = 1/x
y'(1) = 1.
y(1) = e³ since ln 1 = 0.
Equation is y-e^3 = x- 1
y = x + e^3-1.