$\begin{align*} \sum_{i=0}^{\log n}5^{\log n-i}\left(3\cdot2^i+2\right)&=5^{\log n}\sum_{i=0}^{\log n}\left(3\left(\frac25\right)^i+2\left(\frac15\right)^i\right)\\ &=5^{\log n}\left(3\sum_{i=0}^{\log n}\left(\frac25\right)^i+2\sum_{i=0}^{\log n}\left(\frac15\right)^i\right)\;. \end{align*}$
The last two summations are simple geometric series. Assuming that the upper limit of summation is actually $m=\lfloor\log n\rfloor$, they are
$\sum_{i=0}^m\left(\frac25\right)^i=\frac{1-\left(\frac25\right)^{m+1}}{1-\frac25}=\frac53\left(1-\left(\frac25\right)^{m+1}\right)$
and
$\sum_{i=0}^m\left(\frac15\right)^i=\frac{1-\left(\frac15\right)^{m+1}}{1-\frac15}=\frac54\left(1-\left(\frac15\right)^{m+1}\right)\;.$