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it has been years since I have done logs, I remember something like this:

$$x^{\log_z(y)} = y^{\log_z(x)}$$

(where $z$ is the base) Is that correct? It doesn't seem so, since

$$3^{\log_2(4)} \neq 4^{\log_2(3)}$$

Am I right in that assumption? If I am, does


convert to anything easily?

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We have $x^{\log _{z}y}=y^{\log _{z}x}\Longleftrightarrow \left( \log_{z}y\right) \log _{z}x=\left( \log _{z}x\right) \log _{z}y$. I don't understand your inequality because $3^{\log _{2}4}=4^{\log _{2}3}$, $\log _{2}4=2$, $3^{2}=9$ – Américo Tavares Sep 30 '10 at 21:27
so 3log2(4)=4log2(3) ? – naspinski Sep 30 '10 at 21:32
In your notation, yes. – Américo Tavares Sep 30 '10 at 21:39
up vote 5 down vote accepted

We do have

$x^{\log _{z}y}=y^{\log _{z}x}$


$(\log_{z}y)\log_{z}x=(\log_{z}x)\log _{z}y$.

The numerical relation is an equality

$3^{\log _{2}4}=4^{\log _{2}3}=9$.

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If $\rm\:\ \ell\:X\ =\ \log_Z{X}\ \ $ then $\rm\ \ell(X^{\:\ell\:{Y}})\ =\ \ell X\ \ell Y\ =\ \ell Y\ \ell X\ =\ \ell(Y^{\:\ell\: X})$

Or, without taking logarithms $\rm\displaystyle\ \ \: X^{\:\ell\:{Y}}\ \ =\ Z^{\:\ell\: X\ \ell\: Y}\ =\ Z^{\:\ell\: Y\ \ell\: X}\ =\ Y^{\:\ell\: X}$

Your final example is a special case $\rm\ \ Y\ =\ Z^n\:$, $\rm \ $ but it also has a simple direct proof,

namely $\rm\ \ \: X^{\:\ell\: Z^{\:n}}\ =\ X^n\ =\ Z^{\:\ell\: X^n}\ =\ Z^{n\:\ell\: X}\:$.$\:$ Yours is $\rm\: X = 3,\ Z = 2 = n$

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