calculate two-fold difference These are a series of numbers that increase two folds: 
$$0.125, 0.25, 0.5, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024$$
If I pick up two numbers, say $0.5$ and $128$, I want to know know how may $2$-fold difference exist between the two. In this case, there is eight $2$-fold difference between the two numbers.
Is there any formula that the diference can be calculated?
 A: You can find the exponent of the term $r$ as $\log_2 r=\frac{\ln r}{\ln 2}$
If you have two terms $r$ and $s$ which you know are related by doubling you can take $\frac{\ln r-\ln s}{\ln 2}$ where $r\gt s$.
This works for non-integer exponents too.
A: Consider, that $0.5=2^{-1}$ and $128=2^7$. You see, that the numbers have to be expressed by using base 2. Now take a look at the exponents and calculate the difference. Therefore $8\color{blue}{[=7-(-1)]}$ 2-fold differences exist between 128 and 0.5.
Remarks: First, for 2-fold differences you have to transform a given number in a number by using base 2. In case of 0.5 the equation is $0.5=2^x$ The are two opportunities to solve such an equation. 
Firstly using a calculator. 
Take  the log on both sides of the equation. 
$log(0.5)=x\cdot log(2)$. 
Solving for x 
$x=\frac{log(0.5)}{log(2)}$
Take the calculator and input the term on the RHS.
Secondly thinking and calculating by hand (head) 
The question is, how many times you have to divide 1 by 2 to get a result of 0.5 ? Only $\color{red}{\texttt{one}}$ time, because 1 divided by 2 is 0.5. 
And for 128 you have to answer the question, how many times you have to multiply 1 by 2 to get a result of 128 ? Now you do some multiplication operations.
1 multiplied by 2 is 2.
2 multipli2d by 2 is 4.
4 multiplied by 2 is 8.
8 multiplied by 2 is 16.
16 multiplied by 2 is 32.
32 multiplied by 2 is 64.
64 multiplied by 2 is 128.
Therefore 1 has to be multiplied by 2 $\color{red}{\texttt{seven}}$ times.
This method goes much faster, if you know this kind of sequences more or less by heart.
A: Let $T_m$ & $T_n$ be any two randomly selected terms of a geometric progression having first term $a$ & a common ratio $r$ then we have 
$$m_{th}\text{ term},\ T_m=ar^{m-1}$$ $$n_{th}\text{ term},\ T_n=ar^{n-1}$$
Hence the ratio of $m$th & $n$th terms is given as    $$=\frac{T_n}{T_m}=\frac{ar^{n-1}}{ar^{m-1}}=r^{n-1-(m-1)}=r^{n-m}$$
Hence, in general, the $r$-fold difference between any two terms $T_m$ & $T_n$ is given as $$|\text{power of base}\ r \ \text{in ratio}\ r^{n-m}|=\color{blue}{|m-n|}$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{r-fold difference }=|\text{power of base}\  r|=|m-n|}}$$
For the given terms of the above series having common ratio $0.5$ in question, $0.5=2^{-1}$ & $128=2^7$ Here, $m=-1$ & $n=7$ 
Hence 2-fold difference between $0.5=2^{-1}$ & $128=2^7$ is $$|-1-7|=|8|=\color{red}{8}$$
