How to solve for $y$ in the equation $e^y-ey=e^x-ex$?

Question

can I solve for $y$ the equation $$e^y-ey=e^x-ex$$ with a simple trick nice and fast? If not how can I solve this equation with the simplest algebra possible?

Actually I have the function $$f(x)=e^x-ex$$ and I want to find two other functions $g(x)$ and $h(x)$ in order the composition of them equals with the $f(x)$ and if it's possible $g(x)=e^x$

Answer 1

you can use the ProductLogarithmus to solve this equation $$y=\frac{-e W\left(-e^{x-e^{x-1}-1}\right)+e x-e^x}{e}$$

Answer 2

This shows that there exists a parametrization of the solutions.

Exponentiating both sides gives

$$\frac{e^{(e^y)}}{(e^y)^e}=\frac{e^{(e^x)}}{(e^x)^e}$$

Setting $X = e^x$ and $Y = e^y$ we have

$$\frac{e^{Y}}{Y^e}=\frac{e^{X}}{X^e}$$

Let $Y = a X$ then

$$e^{(a-1)X}=a^e$$

so

$$ X = \frac{e \log a}{a-1} $$

or $$ x = \log\frac{e \log a}{a-1} $$

and

$e^y = a e^x$, i.e. $y = x + \log a$ or

$$ y = \log a + \log\frac{e \log a}{a-1} $$

So the solutions $(y(a), x(a))$ are parametrized with $a$. E.g. for $a=2$,

$x = \log\frac{e \log 2}{1} = 0.6335$ and $y = \log 2 +\log\frac{e \log 2}{1} = 1.3266$

and indeed

$$ e^y-ey- (e^x-ex) = e^{1.3266}-e\cdot 1.3266- (e^{0.6335} - e \cdot 0.6335) = 0 $$