Why $\frac{\log4}{\log b}$ can't be simplified to $\frac4b$? 
I want to know why $\frac{\log4}{\log b}$ can't be simplified to $\frac4b$. 

I am a high school student. Please do not quote some theories that are too advanced for me. Thank you!
 A: $\require{cancel}$ It looks like you are thinking of doing this
$$\frac{\cancel{\log}4}{\cancel{\log}b}=\frac4b$$
But don't make the mistake of thinking of "$\log 4$" as "$(\log)\times(4)$". It isn't multiplication (which is what it would need to be if you were to cancel), but it's applying a function. In other words, "$\log 4$" is like "$f(4)$", so $4$ is the input to the function called "$\log$". Unfortunately, it is customary to omit the usual parentheses for functions like $\log$.
If it makes you understand it better, think of it as really meaning
$$\frac{\log(4)}{\log(b)}.$$
A: Assuming you don't have a good professor and something of the sort, remember that $\log$ is the name of an operation. It's like $\cos(x)$. You can't, in general, simplify quotients of functions. So $\frac{\cos30º}{\cos45º}$ is not $\frac{30º}{45º}$. Same thing happens with logarithm.
There are other relations you can use, but you cannot "simplify".
For further information, dividing $\log$ by another $\log$ changes the base of the one above. (I'm not sure if you know this, but when we write $\log$, we usually asume it is logarithm base $e$.) The thing together is:
$$
\frac{\log(A)}{\log(B)}=\log_B(A)
$$
where $\log_B(A)$ is no longer base $e$ (or the original one), but base $B$ instead.
A: You have $\log{4}$ and lets set it equal to $X$.
$$\log{4}=X$$ 
and this has the equivalent exponential form of $10^X=4$.
Similarly, if $\log{b}=Y$, then $10^Y=b$.  Thus,
$$\frac{4}{b}=\frac{10^X}{10^Y}\neq\frac{X}{Y}=\frac{\log{4}}{\log{b}}$$
A: The cancellation rule you know, $$\frac{ab}{ac}=\frac{b}{c}$$ is only true if $ab$ means "$a$ times $b$."
But $\log 4$ doesn't mean multiplying something called "log" by the number $4$. It means we are applying an operation, "logarithm," to the value $4$.  As notation, it looks similar, but it does not mean the same thing as multiplication.
So the rule you know is, verbosely, written as:
$$\frac{a\times b}{a\times c}=\frac{b}{c}$$
When we write $ab$ to mean $a\times b$, then we can cancel like this.
But $\log 4$ does not mean $\log \times 4$, which is meaningless.
Now, as to why $\frac{\log 4}{\log b}\neq \frac{4}{b}$, specifically, the two are equal when $b=2$ or $b=4$, but not for any other $b$. $\frac{\log 4}{\log b}$ is not even defined when $b=1$, since $\log 1 = 0$. 
If $0<b<1$, then $\log b<0$ and thus $\frac{\log 4}{\log b}<0$, but $\frac{4}{6}>0$.
A: The operation $\log\square$ is a black box that eats up the number you write into the square, processes it somehow, and outputs some other number as a result. At the moment you have no big idea what happens in the interior of that box.
There is no reason to assume that for arbitrary $a$ and $b$ one has
$${\log[a]\over\log[b]}={a\over b}\ ,\tag{1}$$
because the inner workings of the black box are much too complicated in order to allow for such a simple "general law".
Imagine for a moment that $(1)$ holds for all $a$ and $b$. Then we would have
$${\log[a]\over a}={\log[b]\over b}\tag{2}$$
for all $a$ and $b$ as well, and this means that ${\log[x]\over x}$ has the same value $c$ for all $x$. This value $c$ could be computed and ISO-certified once and for all. This would further imply that
$$\log[x]= c\>x$$
for all $x$, so that in the end the black box $\log\square$ would not be needed at all. Hm?
A: $$\frac{\log 4}{\log b}=\frac{\log(4^{1/4})^4}{\log(b^{1/b})^b}=\frac{4}{b}\frac{\log\sqrt[4]{4}}{\log\sqrt[b]{b}}$$
so
$$\frac{\log 4}{\log b}\neq \frac{4}{b}\quad{\text{(except when b=4 and b=2})}$$
A: Well, suppose you could do such simplification:
$$
\frac{\log 4}{\log b}=\frac{4}{b}\tag{1}
$$
You would end up with (do you know why?)
$$
b\cdot \log 4=4\cdot\log b,
$$
which implies (do you know why?) that
$$
\log 4^b=\log b^4\tag{2}.
$$
If (1) were true for every $b>0$, then (2) must also be true for every $b>0$ and in particularly true for $b=1$ which is $\log 4=\log 1$. But it is impossible. 
