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:

\begin{matrix} A (n) & = \log (100 n) \\ B (n) & = n^{100} \\ C (n) & = 100^{n} \\ D (n) & = n! . \end{matrix}

Firstly I looked at Archimedes' choice, the logarithmic function.

A (n) = \log (100 n) = \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

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 $2 n + 1$ digits. I then established, where $x$ is the number of digits in $n$ , that $n^{100}$ would have between $100 (x - 1) + 1$ and $100 x$ 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:

$D (n) = n! = \sqrt{2 π n} {(\frac{n}{e})}^{n} .$

As $n$ is a large positive number, $\sqrt{2 π n}$ will be greater than 1. Therefore when $\frac{n}{e}$ is greater than 100, $D (n) > C (n)$ . The 'change over point' occurs for $n$ just less than $n = 100 e$ , 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.

I approximated the function for factorial n further by ignoring the term

\sqrt{2 π n}

\begin{matrix} A (x) & = \log (n) + \log (100) \\ \log (B (n)) & = 100 \log (n) \\ \log C (n) & = n \log (100) \\ \log D (n) & \approx n \log (n) - n \end{matrix} .

I know that

\log (n) < n

for the whole domain of definition of the logarithmic function, and I see that

A

is the smallest. I have to compare the other functions. As

\log (100) \approx 4.60

, so:

\begin{matrix} \log (B) & = 100 \log (n) \\ \log (C) & \approx 4.6 n \\ \log (D) & \approx n (\log (n) - 1) . \end{matrix}

In the limit for large

n

lim_{n \to \infty} \frac{\log n}{n} \to 0

and so

B < C

For large $n$ we can neglect 1 in respect to $\log (n)$ in the approximate formula for $\log (D)$ and, as $4.6 < \log (n)$ we have $D > C > B > A$ .

Here in the diagram, $A (x)$ is represented in red, $B (x)$ in green, $C (x)$ in blue and $D (x)$ in black. It can be observed that for large $n$ (here $n > 300$ ) the order of the magnitude of the functions is $D > C > B > A$ . log graphs