Why is $\log_xy=\frac{\log_zy}{\log_zx}$? Can we prove this using the laws of exponents?
Proof of $\log_xy=\frac{\log_zy}{\log_zx}$
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1You 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. – 2012-12-31
4 Answers
Let $x^a=y$, $z^b=x$ and $z^c=y$. Then $z^{ab}=(z^b)^a=x^a=y=z^c$ so that $ab=c$.
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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0@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 $\log_z (x^{\log_x y}) = \log_z y$. – 2013-01-01
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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0I don't think that second statement is "by definition". That is a _consequence of_ the definition. – 2013-01-03
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! **