For what value of $k$ does $\exp(x)-k=\ln(x)$ have exactly one solution? Is there a closed form for this k?

Question

For what value of k does the equation $e^x-k=\ln(x)$ have exactly one solution? Is there a closed form for this value of k? I know that the numerical value of k is about 2.33036612476, but I don't know if there is a closed-form expression for this number.

Answer 1

There exists $a \in (0,\infty)$ such that $e^{x}-\ln x$ is strictly on decreasing on $(0,a)$ and strictly on increasing on $(a,\infty)$. Also the limiting values at $0$ and $\infty$ are both $\infty$. Hence the given equation has a unique solution iff $k=a$. and we can determine $a$ by setting the derivative equal to $0$.

In other words, $k$ is the unique positive number with $ke^{k}=1$.

Answer 2

You are looking for the zero's of function $$f(x)=e^x-\ln(x)-k$$for which $$f'(x)=e^x-\frac 1x \qquad \text{and} \qquad f''(x)=e^x+\frac 1{x^2} \quad > 0 \quad\forall x$$

The first derivative cancels when $x e^x=1$ and the solution is given in terms of Lambert function, that is to say $\color{red}{x=W(1)}$. At this point, which corresponds to a minimum, there is a double root.

$$f\big(W(1)\big)=e^{W(1)}+W(1)-k \approx 2.33036612476-k$$

Answer 3

Since $\,y=e^x-k\,$ is increasing and concave up, and $\,y=\ln(x)\,$ is increasing and concave down, their graphs must meet tangentially for there to only be a single intersection point. Setting their two derivatives equal to each other, we have that $\;e^x=\frac{1}{x},\;$ i.e. $\,xe^x=1.\,$ Sadly, I don't believe this can be solved analytically.

If we let $\alpha$ be the solution to the above equation, then we have that $\,e^\alpha=\frac{1}{\alpha}\,$ and that $\,\ln(\alpha)=-\alpha,\,$ so that the original equation becomes $\,\frac{1}{\alpha} - k = -\alpha,\,$ which yields $\,k=\frac{1+\alpha^2}{\alpha}.$

So, the solution, not closed-form but I think the best we can do, is

$$k=\frac{1+\alpha^2}{\alpha},\,\text{where}\;\;\alpha e^\alpha=1$$

My trusty graphing calculator shows that $\,\alpha=0.56714329... \,,$ which then yields the value of $k$ that you gave.

Answer 4

One little picture says more than a long speech!

There is only one solution (double-root) $y=k$ for $x=a$ at the minimum of $y(x)$ that is for $y'(a)=e^a-\frac{1}{a}=0$ $$ae^a=1$$ The root of this equation cannot be expressed with a finit number of elementary functions. This requires a special function $W(x)$ called the Lambert-W function: http://mathworld.wolfram.com/LambertW-Function.html

Thus the closed form of the solution is $$a=W(1)\quad\text{also named }\Omega\text{ constant.}$$ $$k=y(\Omega)=\Omega+\frac{1}{\Omega}$$ because $e^\Omega=\frac{1}{\Omega}$ and $\ln(\Omega)=-\Omega$

https://en.wikipedia.org/wiki/Omega_constant

a = Ω = 0.567143290409783872999968662210... (sequence A030178 in the OEIS).

1/Ω = 1.763222834351896710225201776951... (sequence A030797 in the OEIS).

k = Ω+1/Ω = 2.330366124761681...