Comparing $2^{317}$ and $81^{50}$ by hand 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.
 A: Alternative solution:
We have
$$2^{317} > 81^{50} ~ \iff ~ 2^{317\cdot 3/50} > 81^3
~ \iff ~
2^{1/50}2^{19} > 81^3$$
$$~ \Leftarrow ~ \left(1 + \frac{1}{50}\ln 2\right)2^{19} > 81^3 ~ \Leftarrow ~ \left(1 + \frac{1}{50}\cdot \frac{11}{16}\right)2^{19} > 81^3$$
$$~ \iff ~ 811 \cdot 2^{16} > 81^3 \cdot 100
~ \iff ~ (10 + 1/81)\cdot 2^{16} > 81^2 \cdot 100$$
$$ ~ \iff ~ \frac{1}{81}\cdot 2^{16}
> 81^2 \cdot 100 - 10 \cdot 2^{16}$$
$$
~ \iff ~ \frac{1}{81}\cdot 65536
> 6561 \cdot 100 - 10 \cdot 65536 = 740$$
$$~ \iff ~ 65536 > 81 \cdot 740$$
which is true; here, we have used $\mathrm{e}^{y} \ge 1 + y$ for all $y\in \mathbb{R}$, and $\ln 2 > \frac{11}{16}$ which follows from (I learned it from @Jack D'Aurizio)
$$0 < \int_0^1 \frac{x^2(1 - x)^2}{1 + x}\,\mathrm{d} x = \int_0^1
\left(x^3 - 3x^2 + 4x - 4 + \frac{4}{1 + x}\right)\mathrm{d} x = 4\ln 2 - \frac{11}{4}.$$
A: I'm not sure if this is something you consider as 'by hand', but the only thing you really need to calculate to a couple decimals is $2^{1.585}$.
We want to know the ratio $r = \frac{2^{317}}{3^{200}}$. Taking logarithms on both sides $$
\ln r = \ln\left(\frac{2^{317}}{3^{200}}\right) = 317 \ln 2 - 200 \ln 3.$$
Note that $317/200 = 1.585$ and we have that $2^{1.585}$ is slightly bigger than $3$. So in fact we have a lower bound for $\ln r$, hence the ratio by $$
\ln r = 317 \ln 2 - 200 \ln 3 >  317 \ln 2 - 200 \ln 2^{1.585} = 317 \ln 2 - 317 \ln 2 = 0,
$$
so $r > e^0 = 1$, hence $2^{317}$ is slightly bigger.
