Find approximation for size of population over time Assume you start with a population of an objet of size $1$. Assume that a new objet of size $1$ is born at each date and that existing objects double in size in each period. Over time the sequence of populations listed by their sizes will look as follows $(1), (1,2), (1,2,4), (1,2,4,8), (1,2,4,8,16)$, etc. The question I'm having trouble with is: show that the fraction of objects that have size less than $d$ at time $t$ is $\frac{log(d)}{t-1}$ for $d \in \\{1,2,4,...,2^{t-1}\\}$. How to get this expression?

At time $t$ you have your set $\\{1, 2, 4, ... , 2^{t-1}\\}$

Now select some $d \leq 2^{t-1}$.

We must have that $ 2^k < d \leq 2^{k+1}$ for some $k$ in $[0, 1, 2, ... , t-2]$

So now take logs of this inequality to get:

$ k < log(d) < k+1 $

Note that it doesn't matter which base we use as this will just be a multiplicative factor.

So this means that we have k items in the list below the value of log(d) and at time t we have t-1 items. We can see that $k$ is the first integer below $\log(d)$, so as t grows this value becomes ever more accurate.

Hence using this and the fact that there are $t-1$ numbers at time $t$:

Total proportion of numbers less than $d = \frac{\text{total less than d}}{\text{total number at t}} = \frac{\log(d)}{t-1}$