Here are two solutions, one from Simon from Elizabeth College, Guernsey and one from Andrei from Tudor Vianu National College, Bucharest, Romania. First Simon's solution.

The functions are: $$\eqalign { A(n) &= \log(100n) \cr B(n) &= n^{100} \cr C(n) &= 100^n \cr D(n) &= n! .}$$ Firstly I looked at Archimedes' choice, the logarithmic function. $$A(n) = \log(100n) = \log 100 + \log n.$$ I knew that when $n$ gets large $\log 100$ would become irrelevant. This meant that only $\log n$ would be important, and $\log n < n$ when $n > 1$. So $A(n)$ will be less than $n$ for large $n$. We find $n = \log(100)+ \log n$ for $n = 6.472775$ approximately and so $A(n) < n$ for all values of $n> 6.5$.\par The next task was establish whether $B(n)$ or $C(n)$ was greater for large numbers. I was certain of one thing: $B(n) = C(n) = 100^{100}$ for $n=100$.

Firstly I knew that $100^n$ would have $2n+1$ digits. I then established, where $x$ is the number of digits in $n$, that $n^{100}$ would have between $100(x-1)+1$ and $100x$ digits. So $C(n) = 100^n$ is the larger function for $n> 100$.

Therefore, so far, $C(n)$ is the largest function for large values of $n$.

The next task was to compare it to $D(n)$ or factorial $n$.

Using Stirling's formula: $$D(n) = n! = \sqrt{2\pi n} ({n\over e})^n.$$ As $n$ is a large positive number, $\sqrt {2\pi n}$ will be greater than 1. Therefore when ${n\over e}$ is greater than 100, $D(n) > C(n)$. The 'change over point' occurs for $n$ just less than $n = 100e$, that is $ n \approx 270$.

Andrei considered the natural logarithms of the functions and plotted their graphs. For the factorial n function Andrei used Stirling's Approximation which is valid for large n.