I am reading "Problem Solving with Algorithms and Data Structures using Python" and the author is cu

So you have
$\frac{n}{2^i}=1$
Multiplying both sides by $2^i$, one gets
$\Rightarrow <!--\; \Rightarrow \; -->\frac{n}{2^i}\times <!--\; \times \; -->2^i=1\times <!--\; \times \; -->2^i$
You can cancel the $2^i$s on the left hand side, giving us
$n=2^i$
Now - we want some way of turning $2^i$ into just i. How do we do this? Well, applying log base 2 to both sides gives us
$\Rightarrow <!--\; \Rightarrow \; -->{\log }_2<!--\; \; -->(n)={\log }_2<!--\; \; -->(2^i)$
The right hand side, by the definition of the Logarithm is saying "what number do I raise 2 by to get $2^i$?" Clearly, the answer is i. And so
$\Rightarrow <!--\; \Rightarrow \; -->i={\log }_2<!--\; \; -->(n)$
Edit: So technically, the Logarithm in the book should also be base 2; however, I think he may have left it out for 2 potential reasons:
(1) As you say, in many places log without a base specified usually refers to base 10; However, Mathematicians also use log without a specified base to mean base e (where e is Euler's number - don't worry if you don't know what this is) - it could be that the author is just using log without a base to mean base 2 (unlikely, I think - but possible).
(2) I noticed he talked about big O notation. In big O notation, it doesn't really matter whether it's $O({\log }_2<!--\; \; -->(n))$ $O({\log }_{10}<!--\; \; -->(n))$ etc etc. One property of the Logarithm is that
${\log }_a<!--\; \; -->(n)=k\times <!--\; \times \; -->{\log }_b<!--\; \; -->(n)$
That is, the log function in one base can be written as a scalar multiple (just some number) multiplied by the log of another base. And in big O notation
$k\times <!--\; \times \; -->O(f(n))=O(f(n))$
So it doesn't really matter which base you put (provided the base is, of course, positive). Ultimately, the choice is arbitrary in big O notation, and so many times people just write log with the base ommitted. I guess in some sense, you could say there is only really one log function, and the base only really affects what multiple of it you take.