Finding $\Theta$ of Recurrence $ T(n) = 2T(\frac{n}{3})+2T(\frac{2n}{3})+n$ or finding Closed Form for $2^{p+1}+2=3^p$ The goal is to get an exact asymptotic for $T(n).$  I tried two approaches, but both failed: 
1). Recursion tree. We see that $$\sum_{i = 0}^{\log_{3}(n)}n2^i=\Theta(n^{1+\log_3(2)})<<T(n) << \sum_{i = 0}^{\log_{3/2}(n)} n2^i = \Theta(n^{1+\log_{3/2}(2)})$$ but cannot, as I can see, get $\Theta(T(n))$ exactly.
2). Akra-Bazzi Theorem. We get through straightforward calculus that $T(n) = \Theta(n^p)$ where $2+2^{p+1} = 3^p$. As far as I can see this equation yields no closed form for $p\,$ (But it gives a numerical approximation consistent with 1, so that is good).
The goal is to get a closed form better than $2^{p+1}+2 = 3^p.$ I believe such a closed form does exist, for reasons associated with the expected difficulty of the problem, and that it can be found through an alternate derivation, transformations or some other trick.
Summary:
Twin goals of the question, any answer resolving either will be accepted:
(1). Can this problem be solved without the use of the Akra-Bazzi Theorem?
(2). Find a better closed form for $p$.
Essentially, I have a hunch that getting (1) leads right to (2).
Any help is appreciated.
This is  problem 2(m) from Jeffrey Erickson's notes.
 A: The goal to get a closed form better than $\,2+2^{p+1}=3^p\,$ is impossible  because $\,T(n)=An^p-n\,$ is the general solution (and the Akra-Bazzi Theorem is not necessary to get this). 
E.g. the recursion of the Fibonacci numbers is a good example how to get such formulas. The given link above (Jeffrey Erickson's notes) is good too to understand how to solve such recursions.  
For a closed form for $\,p\,$ define a function $\,f(x):=\sqrt[x]{4\cosh x}\,$ for $\,x>0\,$ and $\,f^{-1}(x)\,$ it's inverse function so that one gets:  $$p=\frac{2}{\ln 2}f^{-1}(\sqrt[\ln2]{4.5})$$

Did $\,$gave me the good advice to add a proof for the (descending) monotony of $\,f(x)\,$ .
It's enough to proof this for $\,\ln f(x)\,$ because of $\displaystyle \frac{d}{dx}f(x)=f(x)\frac{d}{dx}\ln f(x)\,$ and $\,f(x)>0\,$ .
We have to proof $\,\displaystyle\frac{d}{dx}\ln f(x)<0\,$ for all $\,x>0\,$ .
$\displaystyle\frac{d}{dx}\ln f(x)=-\frac{\ln (4\cosh x)}{x^2}+\frac{\tanh x}{x}<0\enspace$ is true because of 
$x\tanh x<x<\ln(2 e^x)<\ln(4\cosh x)\enspace$ or in details 
$e^x-e^{-x}<e^x+e^{-x}\,$ , $\enspace x<x+\ln 2\,$ , $\enspace e^x<2\cosh x\,$ .
