A long time ago, I was taught that $\log(ab)=\log a + \log b$ and $\log(a/b)=\log a - \log b$
Then I was reminded of that from this answer on the site: Simple information theory question: where is this equation coming from?
I have no problem applying this rule -- but could someone help explain the intuition behind it so that I truly understand.
Suppose you have a table of powers of 2, which looks like this: (after revision)
$$\begin{array}{rrrrrrrrrr} 0&1&2&3&4&5&6&7&8&9&10\\ 1&2&4&8&16&32&64&128&256&512&1024 \end{array}$$
Each column says how many twos you have to multiply to get the number in that column. For example, if you multiply 5 twos, you get $2\cdot2\cdot2\cdot2\cdot2=32$, which is the number in column 5.
Now suppose you want to multiply two numbers from the bottom row, say $16\cdot 64$. Well, the $16$ is the product of 4 twos, and the $64$ is the product of 6 twos, so when you multiply them together you get a product of 10 twos, which is $1024$.
Notice that to multiply the numbers, we had to count the twos and add the number of twos to get the total number of twos we were multiplying.
The top row is exactly the logarithm of the bottom row, and that is why $\log(ab) = \log a + \log b$. The logarithm of $x$ says how many twos you have to multiply to get $x$. And how many twos must you multiply to get $xy$? You have to multiply all the twos from $x$, which is $\log x$, plus also all the twos from $y$, which is $\log y$, so there are a total of $\log x + \log y$ twos multiplied to make $xy$.
Logarithms are, in a sense, a sophisticated way to count digits/powers of ten (especially if you work in base ten).
Suppose I ask you how many digits $3723\cdot245$ has. You might look at this and reason that $3723$ is on the order of $1000$ (three zeros), and $245$ is on the order of $100$ (two zeros), so we expect $3723\cdot245$ to be on the order of $1000\cdot100=100000$, ($3+2=5$ zeros). And sure enough, $3723\cdot245=912135$, and $912135$ is roughly of order $100000$.
We know how to count digits when multiplying powers of ten; you just add the zeros! We try to extend that reasoning to the rest of the numbers. We might come up with some rule like $d(3723)=3$, and $d(245)=2$, so $d(3723\cdot245)=d(3723)+ d(245)=3+2=5$.
So far, so good.
But wait...
When we try the same thing with $7723\cdot645=4981335$, things start to break down. $d(7723)=3$, and $d(654)=2$, but $d(4981335)=6$. How do we fix this?
Well... Maybe $7723$ is closer to $10000$ than it is to $1000$, and $645$ is kinda halfway between $1000$ and $100$, so let's say $d(7723)=3.5$, and $d(645)=2.5$. So we try $d(7723\cdot645)=d(7723)+d(645)$, and everything seems to work out.
But our reasoning is starting to get sloppy. Let's try to make this digit counting thing more rigorous.
We want to work with powers of ten, because counting digits when you multiply powers of ten is easy. So how many tens are there in $7723$? Well, we just need to solve $7723=10^x$. We get something like $3.8878$. For $645$, we get roughly $2.8096$. For $4981335$, we get $6.6973$. And sure enough, $3.8878+2.8096\approx6.6973$.
Assuming $log$ base 10 for convenience, here is a proof of the first rule: $10^{\log(a)}= a$ and $10^{\log(b)}=b$. So then $ab=10^{\log(a)}10^{\log(b)}=10^{\log(a)+\log(b)}$.
Then take $\log$ base 10 on both sides to get $\log(ab)=\log(a)+\log(b)$.
Similarly you can show the other rule. It might be easier to think of $a/b$ as $ab^{-1}$.
The intuition is that since the logarithm gives you the power needed to raise the base to to get a certain number, it needs to follow the law of exponents.
The rules for logarithms are simply a reflection of the rules for exponents: $$10^x \cdot 10^y = 10^{x+y}.$$ By definition, if we are using the common (base-ten) logarithm, $\log(10^x) = x$, $\log(10^y) = y$, and $\log(10^{x+y}) = x+y$. Using just these equations to substitute for $10^{x+y}$, $x$, and $y$, we can rewrite the equation $\log(10^{x+y}) = x+y$ as $$\log(10^x 10^y) = \log(10^x) + \log(10^y).$$
Alternatively, one can think about logarithms in their historical context. The properties $\log(ab) = \log a + \log b$ and $\log(a/b) = \log a - \log b$ can be viewed as the reason why logarithms were developed in the first place. Multiplying a long list of five-digit numbers by hand is extremely tedious, time-consuming, and error-prone; but armed with a table of five-place logarithms, you can turn that horrid multiplication problem into a problem of merely adding a column of numbers. You can practically define logarithms as "the things that you can add in order to perform multiplication."
And if you really want to develop an ingrained intuition of logarithms, learn how to use a slide rule. If you play around with a slide rule (for example, this one) by moving the sliding rod back and forth, you may notice that whatever number on the D scale is underneath the C scale's number $1$, the number on the D scale underneath the C scale's $2$ is exactly two times the first number. And since the $1$ and $2$ are engraved on the C scale like the marks on a ruler, the distance between them never changes; that is, traveling that distance to the right is the same as multiplying by $2$. Eventually you may understand $\log 2$ as the distance between $1$ and $2$ on your slide rule's C scale, so that each time you move that much further to the right (that is, each time you add $\log 2$ to the distance traveled along the D scale) you double the number on the D scale. If you want the numeric value of that distance, you just have to look below to the L scale, which tells you the logarithm of each number on the D scale; the L scale is marked with numbers that increase as uniformly as the measurements on a ruler.
Based on exponents: $$ x^l = A; log_x(A)=l $$ $$ x^m = B; log_x(B)=m $$ let's say $$ x^n = A B; log_x(AB)=n $$ but also $$ x^n=AB=x^l x^m = x^{l+m} $$ so $$ n=l+m $$ substituting the values we calculated earlier, $$ log_x(AB)=log_x(A)+log_x(B) $$
