Solution of a logarithmic and an exponential curve

Question

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.

Is your equation correct? As orlp stated, the answer would be very trivial then. — M. Winter, May 04 '17 at 15:33

score 2 · Answer 1 · answered May 04 '17 at 15:14

2

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.

answered May 04 '17 at 15:14

orlp

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Solution of a logarithmic and an exponential curve

1 Answers1