Why, or why not, is $5^{\log_3(n)} \in \mathcal{O}(n^2)$ ?
I tried transforming the logarithm to a base of 5, so that the logarithm and power cancel each other out. However, when I try to so I get
$5^{\log_5(n) \cdot \log_3(5)}$
The $\log_3(5)$ obviously is a constant, but since its a factor of the exponent and not a summand, I don't know how to continue.
Any ideas on how to determine the asymptotic behavior of $5^{\log_3(n)}$?
We have
$$5^{\log_3(n)}=\exp\left(\frac{\ln n}{\ln 3}\times\ln 5\right)=n^\alpha$$ where $\alpha=\frac{\ln 5}{\ln 3}$ and since $$\ln 5<\ln 9=2\ln 3\iff\alpha<2$$ then Yes we have
$$5^{\log_3(n)}=n^\alpha=\mathcal O(n^2)$$