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When I look up for properties of the natural Logarithm I found in particular this property

$$ \ln(x^r)=r \ln(x) $$
with $$x\in \Bbb R^{+*} $$ and$$ r\in\Bbb Q$$

My Question is : Why $r$ should be $ \in\Bbb Q$ why not $r\in\Bbb R$

because i can't figure out any problem with being $ r\in\Bbb R$

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  • $\begingroup$ As long as $x>0$, $r$ can be any real number , not only a rational one. $\endgroup$ – Peter Jun 29 '18 at 18:01
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In the terms you put it, there is really no problem: that identity holds for all $r\in\Bbb R$.

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  • $\begingroup$ are you sure Mr Saucy? $\endgroup$ – El Mouden Jun 29 '18 at 17:54
  • $\begingroup$ One hundred percent sure. @ElMouden $\endgroup$ – Saucy O'Path Jun 29 '18 at 17:55
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It should be for all real $r$. The only reason I can see to restrict to $\Bbb Q$ is if you haven't defined $x^r$ for irrational $r$. You can define it for rational $r$ based on the definition for integers and the laws of exponents. Wherever you saw this may not have made the definition for irrational $r$ yet. That usually goes through defining the exponential function or the natural logarithm, with the natural log defined as the integral of $\frac 1x$

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  • $\begingroup$ "The only reason I can see to restrict to $\Bbb Q$ is if you haven't defined $x^r$ for irrational $r$." Or you have defined it for irrational $r$, but proving the property involves quite a bit more machinery than it does for rational $r$. $\endgroup$ – Arthur Jun 29 '18 at 18:45
  • $\begingroup$ i can't understand all of what you said $\endgroup$ – El Mouden Jun 30 '18 at 21:21

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