Write $100$ as sum of $n$ numbers, such that each number is twice as big as its predecessor. I don't quite know where to start on this one.
lets say we have a value 100.
and we want to split it in two parts where one is twice as big as the other.
That would be $v_1 = 66.666$ and $v_2= 33.333$ (sum $100$)
If we want to split the value in 3 parts, where each part is twice as big as the others.
That would be $v_1 = 57.15$ , $v_2 = 28.58$, $v_3 = 14.29 $(sum $100$)
What do I need to do to get $66.66$ from $2$ and $57.15$ from $3$ and so on?
 A: Usually, it helps to formalize the problem. You are looking for an $a$, such that
$$\begin{align}v_1&=a\\ v_2&=2v_1=2a\\ v_3&=2v_2=4a\\ \vdots \\ v_n&=2^{n-1}a\end{align}$$
and $v_1+\dots+v_n=100$, which is the same as 
$$100=a+2a+4a+\dots+2^{n-1}a$$
See, if you can go on from here. The formula for geometric series might help.
A: This has nothing to do with logarithms, but rather to do with simultaneous equations.
In the first case, you want to find two values that sum to 100:
$$x + y = 100.$$
You also have the second condition that one is double the other:
$$x = 2y.$$
You can combine these equations to get
$$ x + y = 2y + y = 3y = 100 \implies y = \frac{100}{3}, x = 2 \cdot \frac{100}{3}.$$
In your second case, you have three numbers:
$$x + y + z = 100,$$
where each number is twice the previous,
$$x = 2y, \\
y = 2z.$$
Again, you can solve these simultaneous equations:
$$x = 2y = 2(2z) = 4z, \\
x + y + z = 4z + 2z + z = 7z = 100 \implies z = \frac{100}{7} = 14.\overline{285714}.$$
A: Arkasim's answer made me realize it can be written as:
split 1 :
1x = 1x = 100
split 2 :
2x + 1x = 3x = 100
split 3 :
4x + 2x + 1x = 7x = 100
split 4 :
8x + 4x + 2x + 1x = 15x = 100
x = 100 / (2^(s-1)*2-1)
where s is the number of splits.
