$L(X) =\exp{\sqrt{(\ln X)(\ln \ln X)}}$, how long does it take to compute $L(X)=L(2^{100})$? I cannot figure how to approach this problem. I could just plug $L(2^{100})$ into Wolfram Alpha and get an answer, but that is not how to actually solve it.
This is the full problem from the book:
Let $L(X) = e^{\sqrt{(\ln X)(\ln \ln X)}}$. Suppose that a computer does one billion operations per second.
(a) How many seconds does it take to perform $L(2^{100})$ operations?
(b) How many hours does it take to perform $L(2^{250})$ operations?
(c) How many days does it take to perform $L(2^{350})$ operations?
(d) How many years does it take to perform $L(2^{500})$ operations?
(e) How many years does it take to perform $L(2^{750})$ operations?
(f) How many years does it take to perform $L(2^{1000})$ operations?
(g) How many years does it take to perform $L(2^{2000})$ operations?
(For simplicity, you may assume that there are $365.25$ days in a year.)
I really have no idea how to even start this problem. Any help would be greatly appreciated.

$L(X)$ is the number of operations to be computed and the computer does $10^9$ $\frac{\text{operations}}{\text{sec}}$ so the time needed to perform $L(X)$ operations is
$$t(X) =  \frac{L(X)~\text{operations}}{10^9\frac{\text{operations}}{\text{sec}}} = \frac{L(X)}{10^9} \text{ sec}$$
Notice that units works like normal fractions so $\frac{\text{operations}}{\frac{\text{operations}}{\text{sec}}} = \text{sec}$ just as $\frac{a}{\frac{a}{b}} = b$ for normal numbers.
To convert to another units we can use that there are 
$$60\frac{\text{sec}}{\text{min}}\cdot 60\frac{\text{min}}{\text{hour}} \cdot 24\frac{\text{hour}}{\text{day}} \cdot 365.25\frac{\text{day}}{\text{year}} = 3.15\cdot 10^7 \frac{\text{sec}}{\text{year}}$$
so the time in years becomes
$$t(X) = \frac{L(X)}{10^9} \text{ sec} \cdot \frac{1}{3.15\cdot 10^{7}\frac{\text{sec}}{\text{year}}} = 3.17\cdot \frac{L(X)}{10^{17}}\text{ year}$$
The same type of conversion can be done for the other units asked for (hours and days).
