Big, bigger, biggest
Which is the biggest and which the smallest of $2000^{2002}, 2001^{2001} \text{and } 2002^{2000}$?
Problem
Which is the biggest and which the smallest of these numbers?
$$2000^{2002} \quad\quad 2001^{2001}\quad\quad 2002^{2000} $$
How do they compare in magnitude?
Student Solutions
Which is the biggest and which the smallest of these numbers and how do they compare in magnitude?
$$A = 2000^{2002},\ B = 2001^{2001},\ C = 2002^{2000}$$
This solution comes from Ilham, St. Patrick's College, Wellington, well done and thank you Ilham.
First let's define the function floor($x$), where $x$ is a real number, such that floor($x$) = the integer part of $x$.
Let $$y = \rm{floor}(\log_a (x)) + 1$$.
As a general rule, y will be the number of digits of $x$ in base $a$. If we reverse this, we can say that $x$ is somewhere between $a ^ y$ and $a^{y + 1}$.
Another basic rule is $\log_a (b^c) = c\log_a (b)$. If we don't use this rule, the calculation cannot be handled using any standard scientific calculators, as they can't handle calculation with numbers greater than $10^{100}$.
If we use these two rules to $A$, $B$ and $C$ in base $10$, it will show that $A$ has $6609$ digits, $B$ has $6606$ digits, and $C$ has $6603$ digits in base 10.
Therefore, $A$ is bigger than $B$ which in turn is bigger than $C$. $A$ is the biggest, and $C$ is the smallest.
A similar solution uses the fact that the logarithm function is an increasing function so it follows that $$\log A > \log B$$ if and only if $A > B$. Hence
$$\log A = 2002 \log 2000 \approx 2002(3.010) \approx 6608.662$$ $$\log B = 2001 \log 2001 \approx 6605.795$$ $$\log C = 2000 \log 2002 \approx 6602.928$$
The approximate difference is given by : $\log A - \log B = \log A/B \approx 3$, hence $A\approx 10^3B$. Similarly $B\approx 10^3C$. Thus $A > B > C.$
Here is Koopa Koo's more general result.
Claim: $A > B > C$
Proof: $A > B$ if and only if $\log A > \log B.$
I shall prove $\log A - \log B > 0$ i.e. $2002\log2000 - 2001\log2001 > 0.$
Let $f(x) = (x + 2)\log x - (x+1)\log(x+1)$ so that for example f(2) = 4log2 - 3log3.
Differentiating this function, $$f'(x) = (x + 2)/x + \log x - 1 - \log(x + 1) = 2/x -\log[(x+1)/x].$$
This derivative is positive if and only if $e^{2/x}> (x+1)/x.$
Using $e^y > 1 + y$ for all $y$, let $y = 2/x$.
We have $e^{2/x}> 1 + 2/x = (x + 2)/x > (x + 1)/x$.
So the function f is increasing, in particular, $f(2000) = \log A - \log B > 0$ and it follows that $A > B$.
The proof that $B > C$ is similar.