Solution
= 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 $\ln x_0=kx_0+1=2$. Hence
$$
\boxed{x_0=e^2,\qquad k=e^{-2}.}
$$
The contact point is $(e^2,2)$, and the resulting <line> $y=e^{-2}x+1$ has exactly the required <derivative> there.