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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

$x^{\log_z(y)}$

convert to anything easily?

  • 1
    In your notation, yes.2010-09-30

2 Answers 2

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We do have

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

because

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

The numerical relation is an equality

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

2

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 $\rm\: X = 3,\ Z = 2 = n$