# Prove that $2 ^ {((\log n)^2) }= n \log n$

I've let $y = 2 ^ {((\log n)^2)}$ and arrive at $\log_2 y = (\log n)(\log n)$, but am unsure how to proceed from here. Help will be appreciated thank you!

• try $n=1$ first. You obtain $2^0=1\cdot0$ i.e. $1=0$. From here you could proceed to prove anything, by a contradiction. – Mirko Apr 26 '18 at 2:40
• Is that log base 2 or some other base. – fleablood Apr 26 '18 at 5:46

$$2^{\log^2(n)}\neq n\cdot\log(n)$$
Substituting $n$ for $1$:
\begin{align} & 2^{\log^2(1)}\neq1\cdot\log(1) \\ \implies& 1\neq0 \\ \end{align}
Assuming $n$ is a natural number and the logs are $2$-based, your expression is wrong: $$2^{(\log n)^2} =2^{(\log n)(\log n)} = (2^{\log n})^{\log n} =n^{\log n} \neq n\log n.$$