seals2001
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(-2)^2 = 4 ?

So why is this undefined ?

Thanks !
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atzy_26
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I thought your question was log-2(4)? Any negative numbers with log (e.g. log-1), the answer is always undefined.

If you are asking why (-2)^2 is undefined on your calculator, then I don't know why.
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RDKGames
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(Original post by seals2001)
(-2)^2 = 4 ?

So why is this undefined ?

Thanks !
y=b^x is not well defined on the real numbers when b<0 so the logarithm isn't either.

You need to introduce complex numbers to deal with this properly.

For instance, we can rewrite

\log_{-2}(4) = 2 \log_{-2}(-2)

and it is easy to be temped to say that \log_{-2}(-2) = 1 ... but actually, this is not the only value it takes.

Using the change of base law we have

\log_{-2}(-2) = \dfrac{\ln(-2)}{\ln(-2)}

but now we can use complex numbers here.

If z = re^{i(\theta+2\pi k)} is a complex number, where k\in\mathbb{Z} and \theta = \mathrm{Arg}(z)\in [0,2\pi) is the principal argument of z, then we have

\ln z =\ln r + i(\theta + 2\pi k)

which means that

\ln (-2) =\ln 2 + i(\pi + 2\pi k)

thus we have

\log_{-2}(-2) = \dfrac{\ln 2 + i(\pi + 2\pi n)}{\ln 2 + i(\pi + 2\pi m)}

for n,m\in\mathbb{Z}.

So, setting n=m gives us \log_{-2}(-2) = 1, but if n\neq m then we have infinitely many solutions.

This is why it is undefined, as there are technically infinitely many values that \log_{-2}(-2) represents.
Last edited by RDKGames; 1 month ago
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seals2001
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(Original post by RDKGames)
y=b^x is not well defined on the real numbers when b<0 so the logarithm isn't either.

You need to introduce complex numbers to deal with this properly.

For instance, we can rewrite

\log_{-2}(4) = 2 \log_{-2}(-2)

and it is easy to be temped to say that \log_{-2}(-2) = 1 ... but actually, this is not the only value it takes.

Using the change of base law we have

\log_{-2}(-2) = \dfrac{\ln(-2)}{\ln(-2)}

but now we can use complex numbers here.

If z = re^{i(\theta+2\pi k)} is a complex number, where k\in\mathbb{Z} and \theta = \mathrm{Arg}(z)\in [0,2\pi) is the principal argument of z, then we have

\ln z =\ln r + i(\theta + 2\pi k)

which means that

\ln (-2) =\ln 2 + i(\pi + 2\pi k)

thus we have

\log_{-2}(-2) = \dfrac{\ln 2 + i(\pi + 2\pi n)}{\ln 2 + i(\pi + 2\pi m)}

for n,m\in\mathbb{Z}.

So, setting n=m gives us \log_{-2}(-2) = 1, but if n\neq m then we have infinitely many solutions.

This is why it is undefined, as there are technically infinitely many values that \log_{-2}(-2) represents.
Amazing explanation !! Thank you
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