smc

We will apply the following logarithmic identity: $${\log }_{p^n}{q}^n={\log }_{p}q,$$ which can be proven by the Change of Base Formula: $${\log }_{p^n}{q}^n=\frac{{\log }_p{q}^n}{{\log }_p{p}^n}=\frac{n{\log }_{p}q}{n}={\log }_{p}q.$$ Now, we simplify the expressions inside the summations: $$\begin{array}{rl}{\log }_{5^k}{3^k}^2& ={\log }_{5^k}(3^k{)}^k\\ & =k{\log }_{5^k}{3}^k\\ & =k{\log }_{5}3,\end{array}$$ and $$\begin{array}{rl}{\log }_{9^k}25^k& ={\log }_{3^{2k}}{5}^{2k}\\ & ={\log }_{3}5.\end{array}$$ Using these results, we evaluate the original expression: $$\begin{array}{rl}\left(\sum _{k=1}^{20}{\log }_{5^k}{3}^{k^2}\right)\cdot \left(\sum _{k=1}^{100}{\log }_{9^k}25^k\right)& =\left(\sum _{k=1}^{20}k{\log }_{5}3\right)\cdot \left(\sum _{k=1}^{100}{\log }_{3}5\right)\\ & =\left({\log }_{5}3\cdot \sum _{k=1}^{20}k\right)\cdot \left({\log }_{3}5\cdot \sum _{k=1}^{100}1\right)\\ & =\left(\sum _{k=1}^{20}k\right)\cdot \left(\sum _{k=1}^{100}1\right)\\ & =\frac{21\cdot 20}{2}\cdot 100\\ & =\boxed{21,000}.\end{array}$$