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. 

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:
$$\log_b x=\frac{\ln x}{\ln b}$$
