# simple/dumb logarithmic conversion inquiry

it has been years given that I have done logs, I bear in mind something similar to this:

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

(where $z$ is the base) Is that proper? It does not appear so, given that

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

Am I right because presumption? If I am, does

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

transform to anything conveniently?

0
2019-05-13 03:46:53
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We do have

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

due to the fact that

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

The mathematical relationship is an equal rights

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

0
2019-05-17 14:41:17
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If $\rm\:\ \ell\:X\ =\ \log_Z{X}\ \$ after that $\rm\ \ell(X^{\:\ell\:{Y}})\ =\ \ell X\ \ell Y\ =\ \ell Y\ \ell X\ =\ \ell(Y^{\:\ell\: X})$

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

Your last instance is a grandfather clause $\rm\ \ Y\ =\ Z^n\:$, $\rm \$ yet it additionally has a straightforward straight evidence,

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

0
2019-05-17 14:38:21
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