I have the equation $$y = e^{y-1}$$and I need to find a way to solve for $y$ algebraically. I already know that the solution is $y=1$, but for the life of me, I cannot find any way to boil the equation down from the original to the solution. Mostly I have tried rearranging the equation and using logarithms, but to no avail.

Please tell me if you know how to solve this specific equation, or know for a fact that this type of equation cannot be simplified.


You know that $$e^x = 1+x+x^2/2 +x^3/6 +....$$

Thus $$e^{(y-1)} = 1+ (y-1)+(y-1)^2/2 +(y-1)^3/6 +....$$

$$= y+(y-1)^2/2+(y-1)^3/6 +... $$ We are looking for a solution to $e^{(y-1)}=y$

Thus we have $(y-1)^2/2+ (y-1)^3/6 +...=0$

which implies $y=1$


Since others have already answered the problem algebraically. To help you with intuition, you can plot the graph of $e^{y-1}$ and $y$ and see what happens.

There is only one intersection point. This is not a proper proof but it helps to imagine things in mathematics.

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