Is $\sqrt{\log (n)}=\frac{1}{\sqrt{2}}*(\log n)$?

Question

Is $$\sqrt{\log (n)}=\frac{1}{\sqrt{2}}*(\log n)$$

Answer 1

No.

Let's interpret $\log$ as the decadic logarithm (base = 10), and let $n = 10000$:

$\sqrt{\log (n)}=\sqrt{\log (10000)}=\sqrt{4}=2$

On the other side:

$\frac{1}{\sqrt{2}}*(\log n)=\frac{1}{\sqrt{2}}*(\log 10000)=\frac{1}{\sqrt{2}}*4=2*\sqrt{2}$

Since $2 \ne 2*\sqrt{2}$, both terms are not equal.

If you prefer to interpret $\log$ as natural logarithm, then let $n = e^4$, and you get the same results.

Answer 2

You are mixing up several identities. The following are true:

• $\log(\sqrt{n}) = \log(n^{\frac{1}{2}}) = \frac{1}{2} \log(n)$
• $\log(n^{\frac{1}{\sqrt{2}}}) = \frac{1}{\sqrt{2}} \log(n)$

But yours isn't true, see e.g. the other answer by @HubertSchölnast.