Find C such that y=lnx+C is tangent to e^x

Question

Given $f(x)=e^x$ and $g(x)=\ln(x)+C$,

find $C\in \mathbb{R}$ such that $f$ and $g$ are tangent to each other.

I tried playing around with a graphing calculator but $C$ doesn't seem to be a rational number. I think I need to differentiate the functions, but not sure how to proceed afterwards.

Answer 1

Suppose there is some point $x_0$ where they are tangent. Then $f(x_0)=g(x_0)$, and also $f'(x_0)=g'(x_0)$. This gives the system of equations for $x_0$ and $C$ as $$e^{x_0}=\ln(x_0)+C$$ $$e^{x_0}=\frac{1}{x_0}$$

The second of these gives $x_0e^{x_0}=1$, which gives the Omega constant $x_0=W(1)=\Omega$ as the solution. So $$C=e^{\Omega}-\ln(\Omega)=\frac{1}{\Omega}+\Omega\approx 2.330.$$

Answer 2

That's a peculiar question and has a peculiar answer! Of course, if the two are tangent at a given x then their derivatives at that x must be the same. That is we must have $\frac{1}{x}= e^x$. That is equivalent to the equation $xe^x= 1$. There is no "elementary" solution to that. We need to use the "Lambert W function" (also called the "omega function" or "product logarithm") which is defined as the inverse function to $f(x)= xe^x$. So, since $xe^x= 1$, $x= W(1)$. In order that the two curves be tangent, in addition to having the same slope, they must touch at that value of x. That is, $ln(W(1))+ C= e^{W(1)}$ so that $C= e^{W(1)}- ln(W(1))$.

Since W(1) satisfies $W(1)e^{W(1)}= 1$ we can write $e^{W(1)}= \frac{1}{W(1)}$. And. taking the logarithm of both sides, $ln(W(1))+ W(1)= 0$ so $ln(W(1))= -W(1)$. So we could also write $C= \frac{1}{W(1)}+ W(1)$. (Thanks to Nathan McDougall for the correction.)

Answer 3

Match the derivatives of the two functions,

$$e^x=\frac 1x$$

which does not have an exact solution in elementary functions. Without resorting to the symbolic Lambert function, an algebraic approximation can be obtained as $x = \frac 13 (1+\ln 2)$. Then, by matching the functional values of the two function, we get,

$$C = e^x - \ln x = \frac 1x + x = \frac{1+\ln 2}{3} + \frac{3}{1+\ln 2}$$

Answer 4

Without resorting to Lambert function,, consider that we look for the zero of function $$f(x)=x e^x-1$$ By inspection or graphing, we can notice that the solution is "close" to $x=\frac 12$.

So, let us build the simplest Padé approximant of $x e^x$; it will be $$f(x) \sim \frac{\frac{1}{2}\sqrt{e}+\frac{13}{12} \sqrt{e} \left(x-\frac{1}{2}\right)}{1-\frac{5}{6} \left(x-\frac{1}{2}\right)}-1$$ and an approximation is then $$x=\frac{34+\sqrt{e}}{20+26 \sqrt{e}}\approx 0.567052$$ while $\Omega=W(1)\approx 0.567143$.

Then $$C=x+\frac 1x=\frac{1556+1108 \sqrt{e}+677 e}{680+904 \sqrt{e}+26 e}\approx 2.33056$$ while $$\frac{1}{\Omega}+\Omega\approx 2.33037$$ showing a relative error of $0.008$%.

Edit

We could even get better approximations building the $[1,n]$ Padé approximant of $f(x)$ itself. Setting the numerator equal to $0$ gives "simple" expressions. For example, $n=2$ would lead to $$x=\frac{148+248 \sqrt{e}+e}{2 \left(28+152 \sqrt{e}+79 e\right)}\approx 0.567144$$