Proof of $\log_xy=\frac{\log_zy}{\log_zx}$

Question

Why is $\log_xy=\frac{\log_zy}{\log_zx}$? Can we prove this using the laws of exponents?

Answer 1

Let $x^a=y$, $z^b=x$ and $z^c=y$. Then $z^{ab}=(z^b)^a=x^a=y=z^c$ so that $ab=c$.

Answer 2

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 $$ z^{\Big((\log_x y)(\log_z x)\Big)} = \Big(z^{\log_z x}\Big)^{\log_x y} = x^{\log_x y} = y. $$

Answer 3

By definition, log_z x^log_xy = log_z y, also by definition we have, log_z x^log_xy = log_xy*log_zx. So log_zy = log_xy*log_zx. With division this gives us the result: log_zy/log_zx = log_xy

Answer 4

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! **