Logarithm calculation result I am carrying out a review of a network protocol, and the author has provided a function to calculate the average steps a message needs to take to traverse a network. 
It is written as $$\log_{2^b}(N)$$
Does the positioning of the $2^b$ pose any significance during calculation? I can't find an answer either way. The reason is, they have provided the results of their calculations and according to their paper, the result would be $1.25$ (given $b= 4$ and $N= 32$).
Another example was given this time $N= 50$, $b=4$ giving a result of $1.41$.
I don't seem to be able to get the same result if I were to apply the calculation and so it's either my method/order of working or their result is incorrect (which I doubt). 
Can someone help to provide the correct way of calculating the values, and confirm the initial results? 
My initial calculation was calculate $\log(2^4) \cdot 32$... Clearly it's totally wrong (maths is not a strong point for me). 

What you want to compute in your given example $N = 32$, $b=4$ is $\log(32)$, in base $2^4 = 16$. 
If you are using a calculator, the "log" function often computes $\log_{10}$ or $\log_e = \ln$ of an argument, so your result is likely computing the log with one of these bases.
You can perform a change of base from $2^b$ to $10$ or $e$ by using the following formula:
$$\log_{2^b} a = {\log_d a \over \log_d 2^b}= {\log_d a \over 2\log_d b}  $$
Where $\log_d$ can be either $d = 10$, or $d = e$, or for your purposes, base $d = 2$.
In the last case, then, where $d = 2$, the given formula simplifies to the equation displayed in Aryabhata nice answer!
