Function that fulfils the equation

Question

Prove that for all $x\in\mathbb{R}$ there exists only one $y=y(x)$ that fulfils the equation: $$3x+e^x=y+e^y$$ I am completely lost with that. What should i do?

Answer 1

Let $$f(x,y)=3x+e^x-y-e^y$$.

We know that in the first place for all fixed real $x$, there always exists at least a real $y$ s.t. $f(x,y)=0$, since $y+e^y$ ranges from $-\infty$ to $+\infty$. So it suffices to show that such a $y$ is unique, which is evident from the fact $$\frac{\partial f}{\partial y}(x,y)=-1-e^y<-1\ne 0$$ due to Implicit Function Theorem.

Answer 2

For any real number $a$, the equation $a=y+e^y$, or $e^y=a-y$, has exactly one solution (think of the plots, and use the Intermediate Value Theorem and the fact that the functions are monotone). So for each $x$, there is a unique $y(x)$ satisfying the equation.