Originally Posted by
HallsofIvy "Doubling" goes 2, $4= 2^2$, $2(4)= 8= 2^3$, etc. In other words, doubling n times gives $2^n$ so you need to solve $2^n= 10000$.
That is an "exponential" problem so take the logarithm of both sides. $log(2^n)= n log(2)= log(10000)$ and then $n= \dfrac{log(10000)}{log(2)}$.
That is the "number of doublings" and since each doubling requires 2.4 days, divide that by 2.4 to get the number of days.
You could also do that just by continuing: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384 which is larger than 10000, but that won't give you the fraction between 13 and 14 doublings.