Given $f(x)= e^x - e^ax$ with roots $P$ and $Q$,$0I have a midterm tomorrow and while I was looking through old exams from my professor I stumbled on a problem for which I'm not able to see the solution.
We want to find the rots of $f(x) = e^x - xe^a$ with $a>1$.
Consider the fixed point functions $g_1(x) = e^x/e^a$ and $g_2(x) = a + \ln(x)$.
First, I had to show that $f(x)$ has two root $P$ and $Q$ such as $0<P<1<a<Q$ which I did using the Intermediate value theorem and the fact that $g_1(x)$ and $g_2(x)$ are strictly increasing.
My problem is this:
(a) Show that $g_1(x)$ and $g_2(x)$ have exactly two fixed points each and they coincide with the roots of $f(x)$. 
(b) Then show that $g_1(x)$ doesn't converge to $Q$ and $g_2(x)$ doesn't converge to $P$. 
I tried to show (a) by setting $g_1(x) = e^x/e^a = x$ and $g_2(x) = a + \ln(x) = x$ but I got stuck.
I also tried arguing that if: 


*

*$g_1(x) \in C[0,1]$ and $g_1(x) \in [0,1]$  $\forall x \in [0,1]$

*$g_1'(x) \in C[0,1]$ and $\exists K$ $0<K<1$ s.a $|g_1'(x)| \leq K$ in $[0,1]$


Then there id a unique fixed point in $[0,1]$ and $x_{n+1} = g_1(x_n)$ converges to $P$
And the same argument for an interval $[a,a+1]$ so that there would be two unique fixed point, but the conditions don't hold for that interval.
Some help would be greatly appreciated, I really am stuck on this problem and the midterm I'm preparing for is tomorrow afternoon.

Notice that 
$$f(P)=0\Rightarrow e^P-e^aP=0\Rightarrow e^P=e^aP\Rightarrow P=\frac{e^P}{e^a}=g_1(P)$$
In the same way for $Q$. So $g_1(x)$ has two fixed points.
Now 
$$f(P)=0\Rightarrow e^P-e^aP=0\Rightarrow e^P=e^aP\Rightarrow P=a+\ln(P)=g_2(P)$$
The same technique for $Q$.
