# Solve $\ln \,\ln \big(n^{\frac{1}{2^{k}}}\big)=\ln\,\ln\,(n)-k\,\ln 2$.

## Question

While solving a Recurrence Relation, I got stuck at some Logarithmic equation.For this i have to prove

$$\ln \,\ln (n^{\frac{1}{2^{k}}})=\ln\,\ln\,(n)-k\,\ln 2$$

## My Approach

$$\ln \,\ln (n^{\frac{1}{2^{k}}})=\frac{1}{2^{k}}*\ln \,\ln(n)$$

if i convert $2^{k}$ to $k\,\ln\,2$ ,then

$$\ln \,\ln (n^{\frac{1}{2^{k}}})=\frac{\ln \,\ln(n)}{k\,\ln\,2}$$ $$=\ln\,\ln\,(n)-k\,\ln 2$$

but i am not getting how to get

$2^{k}$ to $k\,\ln\,2$ Please help me out!

$\ln \,\ln (n^{\frac{1}{2^{k}}})=\ln [(\frac{1}{2^{k}})\ln(n)] = \ln({2^{-k}}) + \ln(\ln n) = -k \ln2 + \ln(\ln n)$
You can't turn $2^k$ into $k\ln 2$ since it is not true, for example, if $k = 1$, then $2^k = 2 \neq \ln 2 = k\ln 2$.
Your mistake is $$\ln\ln(n^{1/2^k})= \frac 1{2^k}\ln\ln n.$$
You can't ignore that there are two logarithms here and $\ln\ln$ doesn't follow the same rules as $\ln$. What you can do is what ab123 did in their answer:
$$\ln\ln(n^{1/2^k})= \ln\left(\frac 1{2^k}\ln n\right)$$ and then use that $\ln(xy) = \ln x + \ln y$.