simple question: why does determining the log help to find the number of digits in a number? For example, i see questions like $125^{100}$ or $8^{25}$ and they ask to find how many digits are in the number. why does taking the log give you this? i know how to actually calculate it, i'm just confused about why. 
 A: Any number $n\in\mathbb{N}$ is between $10^k \leq n < 10^{k+1}$ for some $k\in\mathbb{N}$. This means that the number of digits of $n$ is $k+1$. When you take $\log_{10}$ of all sides, you get $k \leq \log{n} < k+1$, which basically gives you $k+1$ if you take the floor of $\log{n}$ and add $1$, i.e., $\lfloor\log{n}\rfloor+1$.
A: Since you say you know how to calculate this already and that you just want a better understanding of why it works, let me attempt to give you some intuition.
$\log_b{n}$ gives the answer to the question ‘how many times must we multiply $b$ by itself in order to reach $n$?
When we write a number in base $b$, each digit corresponds to a power of that base. So, for example in base $10$ we can write $$362 = 3 \cdot 10^2 + 6 \cdot 10^1 + 2 \cdot 10^0$$
We can see that each power in this sum is one less than the index of the digit, counting from the right. So the number of digits will be equal to the highest power, plus one. What we need to calculate is the smallest power of $10$ such that we overshoot $362$.
$log_{10}{362} \approx 2.56$ and so we only reach $362$ after multiplying twice ‘and a bit more’. So, since we want an integer number of digits we must multiply $3$ times.
Formally, $\lceil \log_b{n} \rceil$ gives you the number of digits of $n$ in base $b$. So for example, the number of digits of $52373598$ in base $10$ is given by $\lceil \log_{10} 52373598 \rceil= 8$
