Doubt with a problem of grown functions and recursion tree
Instead of having logarithms in the working, I'll substitute $n=2^m$:
$$
T(n)
= n\sum_{i=0}^{m} \left(\frac{3}{2}\right)^i
= n \frac{\left(\frac{3}{2}\right)^{m+1}-1}{\frac{3}{2}-1} 
= 2n \left(\left(\frac{3}{2}\right)^{m+1}-1\right)
= 2n \left(\frac{3}{2}\right)^{m+1}-2n
$$
I believe the previous step is where you didn't expand the $2n$ into the first term. I am also not making any guess of what the master theorem or order of complexity is. Guesses should not be necessary anyways, since we just assume $T(1) \in \Theta(1)$.

Simplifying by keeping only the first term, and removing constants:

$$T(n) \in \Theta(n 1.5^{\log{n}})$$

Then, with some rearranging:

$$n 1.5^{\log_2{n}} = 2^{\log_2{n}}1.5^{\log_2{n}} = 3^{\log_2{n}}=2^{\log_2{3}\log_2{n}}=n^{\log_2{3}}$$

Thus:

$$T(n) \in \Theta(n^{\log_2{3}})$$