# 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?

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What you wrote is not true, if $X\neq x$. Perhaps you mean to ask why $\log_x Y=\frac{\log_X Y}{\log_X x}$, or why $\log_x x=1$. –  Eric Naslund Dec 31 '12 at 20:57
You seem to be using lower-case $x$ and capital $X$ to refer to the same thing. That shouldn't be done. In standard usage, mathematical notation is case-sensitive. –  Michael Hardy Dec 31 '12 at 20:58

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

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

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Also $\log_{z}x^\log_{x}y = \log_{z}y$ which you can then simplify to $y=y$ (And could someone edit this, because I am not understanding the programming) –  Alex Dec 31 '12 at 21:53
@Alex : You can edit your own comment within five minutes of posting it, but not after that, and others can't edit it. It seems you mean $\log_z (x^{\log_x y}) = \log_z y$. –  Michael Hardy Jan 1 '13 at 0:37

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

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I don't think that second statement is "by definition". That is a consequence of the definition. –  Michael Hardy Jan 3 '13 at 0:46

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! **

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