Solving for an exponent using logarithms

Question

I'm looking at the solutions to one of my homework assignment, which states that if ${n^{1/2}}^h = 2$, then $h = log(log(n))$ (we are trying to solve for h). How did they arrive at this conclusion?

Answer 1

$n^{1/2^h}=2$

$(\frac{1}{2})^h=\frac{1}{2^{h}}=2^{-h}$

$n^{1/2^h}=n^{2^{-h}}$

taking logarithm in base 2 we get:

$2^{-h} \log _2 n=1$

taking logarithm in base 2 we get:

$- h \log_2 2+\log_2(\log_2 n)=0$

$-h+\log_2(\log_2 n)=0 $

⇒ $h = \log_2(\log_2 n)$

Answer 2

It's important to note that these $\log$'s are base $2$. Take $\log$'s of both sides and pull out the exponent:

$$\left(\frac{1}{2}\right)^h \log n = \log 2 = 1.$$

Multiply through by $2^h$ to get:

$$\log n = 2^h.$$

Take $\log$'s of both sides again:

$$\log\log n = h\log 2 = h.$$