How to compare these two numbers without calculator:
$2^{317}$ and $81^{50}$
(Pen & paper test)
I thought about using logarithms and doing Taylor approximation, but these numbers are close to one another and I'd need a lot of Taylor expansion summands which defeats the purpose as raising 3 to some power of 15 and operating those fractions is not something you'd do by hand.
I've seen similar questions but there the powers were "nice" and are possible to simplify / reduce, yet there I don't see such an opportunity.
Best Answer
We have $$ 2^{317} > 81^{50} \iff 2^{17} > \left(\frac98\right)^{100} \iff \log_e2 > \frac{100}{17}\log_e\frac98, $$ and $$ \log_e2 = \log_e\frac{1 + 1/3}{1 - 1/3} > 2\left(\frac13 + \frac1{3\cdot3^3} + \frac1{5\cdot3^5}\right) = \frac23\left(1 + \frac1{27} + \frac1{405}\right) = \frac{842}{1215}, $$ and \begin{gather*} \frac{100}{17}\log_e\frac98 = \frac{100}{17}\log_e\frac{1 + 1/17}{1 - 1/17} = \frac{200}{17^2}\left(1 + \frac1{3\cdot17^2} + \frac1{5\cdot17^4} + \cdots\right) \\ < \frac{200}{17^2}\left(1 + \frac1{3\cdot17^2}\left(1 + \frac1{17} + \frac1{17^2} + \cdots\right)\right) = \frac{200}{17^2}\left(1 + \frac1{3\cdot16\cdot17}\right) \\ = \frac{200\cdot817}{3\cdot16\cdot17^3} = \frac{25\cdot817}{6\cdot4913} = \frac{25\cdot817}{29478} < \frac{25\cdot817}{29475} = \frac{817}{1179}, \end{gather*} and $$ 842\cdot1179 = 992718 > 992655 = 1215\cdot817, $$ therefore $$ \log_e2 > \frac{842}{1215} > \frac{817}{1179} > \frac{100}{17}\log_e\frac98, $$ therefore $$ 2^{317} > 81^{50}. $$