 Solution

Let the contact point have positive horizontal coordinate $x_0$. Equality of slopes between the tangent line and the logarithmic curve requires $k=1/x_0$. The point must also satisfy both equations, giving $\ln x_0=k{x}_0+1=2$. HenceThe contact point is $(e^2,2)$, and the resulting line $y=e^{-2}x+1$ has exactly the required derivative there.