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We know that $\log_a xy=\log_a x + \log_a y$ and that logarithms of negative numbers are undefined. But what happens if we try to apply this property to let's say $-5$?

$\log_a-1*5=\log_a-1+\log_a5$

After that we can infinitely apply the same property, but where do we actually get the undefined part?

molarmass
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Michael
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    The log of a negative number is defined, albeit, non-uniquely. For $x<0$, we have $\log x=\log|x|+i(2n+1)\pi$ for integer values of $n$. – Mark Viola Dec 06 '15 at 19:35
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    You have written $\log_a(-1\cdot 5) = \log_a(-1) + \log_a(5)$. Note that the term $\log_a(-1)$ on the right hand side is not defined. – Mankind Dec 06 '15 at 19:36
  • Maybe I am thinking this programmatically but similarly the $log_a (-1)$ would give $log_a(-1) + log_a1=log_a(-1)$, therefore the computation wouldn't halt somewhere, giving a clear undefined result. – Michael Dec 06 '15 at 19:48
  • By that logic $1=1+0$, and thus the computation wouldn't halt. Therefore $1$ is undefined? – YoTengoUnLCD Dec 06 '15 at 20:16
  • Such challenges leads to creating new number fields. Dr. MV provided with the insight - expanding the real numbers into complex numbers allow dealing with the challenge presented. – Moti Dec 06 '15 at 21:22

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