What is the least positive integral value of $k$ such that the equation $$\ln x + k = e^{x-k}$$ has a solution?
I tried plotting the two curves and concluded that the limiting value of k will be the case when $y = \ln x +k$ and $y = e^{x-k}$ touch each other. But then I could not proceed further to find the solution.
This answer doesn't provide a lot of insight into the why, but for $k=x=1$ we have:
$$\ln 1 + 1 = e^{1-1}$$ $$0 + 1 = 1$$
Since $k =1$ is the least positive integer anyways, this solution must be optimal.