As far as I know for $\log_2 x$ to be defined $x$ must be higher than 0. However when I enter $2^{\log_2(-5)}$ into wolframalpha it gives result $-5$. Is it mistake?
Is $2^{\log_2(-5)}$ defined?
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$\begingroup$
logarithms
wolfram-alpha
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0I just entered 2^(log_2(0)) into Wolfram Alpha and it gave me $0$. Then I tried 1^(log_1(2)) and it said it's indeterminate. Then I tried log_1(2) and it said $\tilde{\infty}$, i.e. $\infty$ with a tilde over it. – 2012-05-28
2 Answers
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Yes...and, no.
There are ways to define $\log_2(-5)$. They require knowing something about complex numbers. And if you use one of those ways to define $\log_2(-5)$, then $2^{\log_2(-5)}=-5$.
Now, if WA says $2^{\log_2(0)}=0$, then I'll be worried.
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0WA uses the change of basis: $\,\displaystyle{\log_2x=\frac{\ln x}{\ln 2}}\,$ and then they call $\ln$ the natural logarithm... – 2012-05-28
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It is not a mistake, though you are right to be worried. What they are using is the complex logarithm. It is possible to define logarithms $\log z$ of any nonzero complex number $z$, just by saying that $\log z$ is a complex number $w$ with the property that $e^{w} = z$. Unfortunately there are infinitely many such solutions, so there are infinitely many logarithms $\log z$ of $z$. By definition, though, they all have the property that $e^{\log z} = z$, which is what wolframalpha gave you.
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0With slight modifications to deal with base-2 logarithms rather than the natural logarithm that froggie is writing about. – 2012-05-28