Vessels received by the brother monarch 
An eastern monarch sends 10.000 golden vessels to a brother monarch, whose kingdom is many days march distant. The gift is carried on camels. Each merchant who supplies camels for some part of the journey, demands as commissions 10% of what passes through his hands. Thus the first merchant hands over to the second 9.000 golden vessels. Altogether the vessels pass through 20 merchants. How many vessels does the brother monarch receive at the end? -- W.W Sawyer "Mathematician's Delight"

The time consuming solution is to manually calculate all the values, but there must be a logarithmic rule/formula that can used to find the result quickly for any number of merchants (without writing a computer program).
 A: Each time the vessels pass through a merchant, their number is multiplied by $.9$.  Thus after $20$ merchants, the number of vessels is multiplied by $.9^{20} \approx 0.12158$.  So in the end you have $10000(0.12157665459) \approx 1215.8$ vessels.
A: Let $n_k$ be the number of vessels after passing through $k \geq 0$ merchants. We thus have
$$n_{k+1} = 0.9 n_k$$
with initial condition $n_0 = 10^4$. The general solution is thus $n_k = 0.9^k n_0$. If you want to take into account the fact that $n_k$ should be an integer, you can use the floor function $\lfloor \cdot \rfloor$, i.e.,
$$n_{k+1} = \lfloor 0.9 n_k \rfloor$$
Here's an experiment in Haskell (using the GHCi interpreter):
Prelude> let f n = floor (0.9 * fromIntegral n)
Prelude> let n0 = 10^4
Prelude> let ns = iterate f n0
Prelude> take 21 ns
[10000,9000,8100,7290,6561,5904,5313,4781,4302,3871,3483,3134,2820,2538,2284,2055,1849,1664,1497,1347,1212]

Hence, after $20$ merchants there will only be $1212$ vessels.
A: It seems you need 
$
\begin{split}
10000*0.9^{20}
  &= 10000*(1-0.1)^{20} \\
  &= 10000*(1 - 20*0.1 + \frac{20*19}{2} * 0.01 - \frac{20*19*18}{6}*0.001 + \frac{20*19*18*17}{24}*.0001 - \frac{20*19*18*17*16}{120}*.00001) \\
  &\approx 10000*(1 - 2 + 1.9 - 1.14 + 0.4845 - 0.1550)\\
  &= 1295
\end{split}$
feel free to add some terms if you need
A: This isn’t so much an answer (though it does contain a good estimate) as a brief visit to the past.
Having grown up before calculators, and having memorized the $5$-place decimal logs of $2,3$, and $7$ over $50$ years ago, I was curious to see how close I could come using only easy hand computation. We want $10000(0.9)^{20}$, whose log is $4+20\log0.9\approx4+20(0.95424-1)=3.0848$, so it’s going to be a bit over $1000$.
$\log 5-\log 4\approx 0.69897-0.60206=0.09691$ and $\log6-\log5\approx0.77815-0.69897=0.07918$, so it’s between $1200$ and $1250$, closer to the former. After this it gets harder, but $1200=5\cdot240$, $1250=5\cdot250$, and by good fortune I recognize that interval as containing $243=3^5$, so we might try $3^5\cdot5=1215$: $\log1215=5\log3+\log5\approx 5\cdot0.47712+0.69897=3.08457$, so $1215$ is pretty close. (Hey, it beats counting sheep!)
