Comparing powers

Question

Is there any way to check besides acctually calculating whether $n^x$ is $> = <$ then $m^y$?

For example how to check wheter $440902^{532446} > = < 555151^{523163}$?

Answer 1

You can make the calculation easier (or at least put it on a more eyeball-able scale) by using logarithms or other monotonic functions. Basically, you want to use a function $f$ such that $x > y$ and $f(x) > f(y)$ are equivalent.

When dealing with exponentiation, logarithms are particularly useful since you can do a lot of manipulations with them. For example:

$\log{(440902^{532446})} = 532446\log{440902}$ (which is true regardless of the base of the log)

Or possibly even more useful:

$\log_{10}{440902} = \log_{10}{100000} + \log_{10}{4.40902} = 5 + \log_{10}{4.40902}$

So the log, base 10, of the left-hand side of your question is $\approx 532446(5 + \log_{10}{4})$ while the log of the right-hand side is $\approx 523163(5 + \log_{10}{5})$. Since the logs of 4 and 5 are both going to be about 3-ish, the left-hand side has got an obvious lead.

Answer 2

Intuitively, the exponent matters much more than the base, so without computation I would guess the left side is greater. Alpha confirms this. The log of the left is over $100$ more than the log of the right, so the left is greater than the right by a factor over $e^{100} \approx 2.7\cdot 10^{43}$. To be surer, take the log of each side.