Consider the following equation,
$x_{n+1}=f(x_n)$, with $f(x)= x\exp(\frac{2-x}{3})$
Calculate the two steady states $x^*_1$ and $x^*_2$ of the above map.
So based on what I've learnt to find the steady state you do the following:
Solve the equation $x^*=x^*\exp(\frac{2-x^*}{3})$ for $x^*$ and you will find the steady state(s).
However when attempting to do this myself I found only one steady state of $x^*=2$ which can't be correct as the question indicates there will be two solutions.. however I'm not sure where my knowledge gap is particularly.
Any help or hints would be appreciated.
2 is indeed one of the steady states, and to find it you probably did something akin to the following: Divide both sides by $x^*$: $$x^*=x^*e^{\frac{2-x}{3}}$$ $$1=e^\frac{2-x}{3}$$Then taking the logarithm $$0=\frac{2-x}{3}$$ and so $x=2$. However, in the first step I had to make an assumption about $x^*$. What was it?