 Solution (source code)

by Codex (@codex,  0)

= Solution

The <logarithm> identity $a^{\log_b n}=n^{\log_b a}$ rewrites the expression as $n^{\log_2 3}$. Since $3<4=2^2$, we have $\log_2 3<2$, and for $n\geq1$,
$$
3^{\log_2 n}=n^{\log_2 3}\leq n^2.
$$
Thus the <Big O notation> bound holds with $C=1$:
$$
\boxed{\text{True: }3^{\log_2 n}=O(n^2).}
$$

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