Expressing logarithms as ratios of natural logarithms

$$\frac{\log_2 x}{\log_3 x}=\frac{\ln x}{\ln 2} \div \frac{\ln x}{\ln3}$$

Why can logarithms be written as ratios of natural logarithms?
Can you explain it abstractly, please?
Example of an abstract explanation: the logarithm function is an isomorphism from the group of positive real numbers under multiplication to the group of real numbers under addition, represented as a function.

The teacher doesn't go into abstractions in class, so I would really like to understand it in an abstract sense. Thank you.

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I'm gathering that you want something like a proof that every such isomorphism can be expressed as such a ratio? – dfeuer Oct 29 '13 at 2:53
@dfeuer that would be interesting, good idea. – Emi Matro Oct 29 '13 at 3:00
Bad mathematics alert! How are group or isomorphism abstractions and $x$, $\ln$, $2$, $3$, not? The simplest and most elementary explanation is the best you can get of this. Being sesquipedalian doesn't make the explanation more or less abstract, or better. – Mlazhinka Shung Gronzalez LeWy Oct 29 '13 at 3:12
user4150, as lhf notes in a comment below, this is essentially impossible unless you bring in properties of the function that aren't really algebraic in nature. – dfeuer Oct 29 '13 at 3:17
@dfeuer I see, thanks. – Emi Matro Oct 29 '13 at 3:20

Consider $y=\log_b x$. Then, by definition, $b^y=x$ and so $y \ln b = \ln x$. Thus, $$\log_b x=\frac{\ln x}{\ln b}$$
A more sophisticated argument is the following. Consider a continuous function $L:\mathbb R^+ \to \mathbb R$ such that $L(xy)=L(x)+L(y)$. Then $F(x)=L(e^x)$ satisfies $F(x+y)=F(x)+F(y)$ and is thus a continuous automorphism of the additive group $\mathbb R$. It is easy to see that $F$ must be a scaling: $F(x)=ax$ for some $a$. Of course, $a=F(1)=L(e)$. When $L(x)=\log_b x$, we have $a=\log_b e=\dfrac1{\ln b}$. Since $L(x)=F(\ln x)$, we have, as before: $$\log_b x=\frac{\ln x}{\ln b}$$
I gather from your answer that there are probably other isomorphisms between multiplicative $\Bbb R_+$ and additive $\Bbb R$ that I imagine are horrible relatives of "additive monsters", and thus likely nowhere continuous, nowhere locally bounded, etc.? – dfeuer Oct 29 '13 at 3:08