Proof of $\log_xy=\frac{\log_zy}{\log_zx}$ Why is $\log_xy=\frac{\log_zy}{\log_zx}$? Can we prove this using the laws of exponents?
 A: I will presume that what was meant was $\displaystyle\log_x y = \frac{\log_z y}{\log_z x}$.
Notice that this is true if and only if $(\log_x y)(\log_z x) = \log_z y$, and that holds if and only if $\displaystyle z^{(\log_x y)(\log_z x)}=y$.
So
$$
z^{\Big((\log_x y)(\log_z x)\Big)} = \Big(z^{\log_z x}\Big)^{\log_x y} = x^{\log_x y} = y.
$$
A: By definition,
logz xlogxy = logz y, also by definition we have, logz xlogxy = logxy*logzx. So logzy = logxy*logzx. With division this gives us the result: logzy/logzx = logxy
A: Let $x^a=y$, $z^b=x$ and $z^c=y$. Then $z^{ab}=(z^b)^a=x^a=y=z^c$ so that $ab=c$.
A: Demonstrate: $ \log_xy=\frac{\log_ay}{\log_ax}; x,y,a \in \mathbb{R} $
We initially have:$$ f(x,y)= \log_xy$$ 
We transform it to the exponential form: $$ x^{f(x,y)}=y$$
We apply logarithm of base $a$ for $a \in \mathbb{R}$ on both sides of the equation:$$\log_a{x^{f(x,y)}}=\log_ay$$
Applying the exponential property: $\log_ab^c=c\log_ab$ we have: $$ f(x,y) \log_ax=\log_ay$$
Getting $f(x,y)$ $$ f(x,y)=\frac{\log_ay}{\log_ax} $$
We initially have that $f(x,y)=\log_xy$ and then we got $f(x)=\frac{\log_ay}{\log_ax}$, so:
$$ \log_xy = \frac{\log_ay}{\log_ax} \\ Q.E.D.$$
** If there are words that are not appropriated in this demonstration please let me know. I'm not English speaker!  **
