Common Point

by swapping x,y, its easy to prove that they are inverse function of each other.

this means one is a reflection of other w.r.t y=x.

since they share a common point, intuitively they must share a common tangent y=x.

1. find the point which gradient is 1 (same gradient as y=x). $\begin{aligned} y &= \frac{1}{a}e^x \\ \frac{dy}{dx} &= \frac{1}{a}e^x = 1 \\ x &= \ln a \\ y &= x = \ln a \\ \text{tangent point } &= (\ln a, \ln a) \end{aligned}$

2. use the tangent point to find a. $\begin{aligned} \text{subs } ( \ln a, \ln a ) \text{ into } y &= \frac{1}{a}e^x \\ \ln a &= \frac{1}{a}e^{\ln a} \\ \ln a &= 1 \\ \therefore a &= e \end{aligned}$

When two curves $y_1(x)$ and $y_2(x)$ share only one common point $x_0$ , then $y_1(x_0) = y_2(x_0)$ and $y'_1(x_0) = y'_2(x_0)$ . Therefore, we have:

$\begin{cases} y_1(x_0) = y_2(x_0) & \implies \dfrac {e^{x_0}}a = \ln a + \ln x_0 & ...(1) \\ y'_1(x_0) = y'_2(x_0) & \implies \dfrac {e^{x_0}}a = \dfrac 1{x_0} \implies a = x_0 e^{x_0} & ...(2) \end{cases}$

Putting $a = x_0 e^{x_0}$ in $(1)$ ,

$\begin{aligned} \frac {e^{x_0}}{x_0 e^{x_0}} & = \ln (x_0 e^{x_0}) + \ln x_0 \\ \implies \dfrac 1{x_0} & = 2 \ln x_0 + x_0 \end{aligned}$

We note that $x_0 = 1$ is the solution. Therefore $a=x_0 e^{x_0} = e \approx \boxed{2.718}$ .

There is a way to do this without using the Intuition and calculus at all and it involves the lambert w function.

W = f^{-1} where f(x) = xe^x Domain(W) = [-1/e , inf) However W is bivalued on (-1/e , 0) The graph is given by x=ye^y where W(-1/e)=-1

Now the two functions in the question are inverse functions. Hence any intersection between the two happens on the line y=x

Let the intersection point be (t,t) This implies

at=e^t te^{-t}=1/a -te^{-t}=-1/a

Taking W on both sides

W(-te^{-t})=W(-1/a) -t=W(-1/a) t=-W(-1/a)

Since the curves intersect exactly once, t must have exactly 1 solution. This is to say that -1/a must be in the single valued domain of the Lambert W function.

Therefore -1/a belongs to [-1/e,inf) And -1/a does not belong to (-1/e,0) Hence -1/a belongs to {-1/e}U[0,inf)

Case 1 -1/a belongs to [0,inf) -1/a >= 0 1/a <= 0 a <= 0 However this is not possible as a>0

Case 2 -1/a belongs to {-1/e} -1/a=-1/e a=e t=-W(-1/e)=-(-1)=1

Hence the point of intersection is (1,1) and a=e