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Why is log-2(4) not 2 ?

(-2)^2 = 4 ?

So why is this undefined ?

Thanks ! :smile:
Reply 1
Avatar for atzy_26
atzy_26
10
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.
Original post by seals2001
(-2)^2 = 4 ?

So why is this undefined ?

Thanks ! :smile:

y=bxy=b^x is not well defined on the real numbers when b<0b<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)=2log⁡−2(−2)\log_{-2}(4) = 2 \log_{-2}(-2)

and it is easy to be temped to say that log⁡−2(−2)=1\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)=ln⁡(−2)ln⁡(−2)\log_{-2}(-2) = \dfrac{\ln(-2)}{\ln(-2)}

but now we can use complex numbers here.

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

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

which means that

ln⁡(−2)=ln⁡2+i(π+2πk)\ln (-2) =\ln 2 + i(\pi + 2\pi k)

thus we have

log⁡−2(−2)=ln⁡2+i(π+2πn)ln⁡2+i(π+2πm)\log_{-2}(-2) = \dfrac{\ln 2 + i(\pi + 2\pi n)}{\ln 2 + i(\pi + 2\pi m)}

for n,m∈Zn,m\in\mathbb{Z}.

So, setting n=mn=m gives us log⁡−2(−2)=1\log_{-2}(-2) = 1, but if n≠mn\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)\log_{-2}(-2) represents.
(edited 3 years ago)
Reply 3
Avatar for seals2001
seals2001
OP
16
Original post by RDKGames
y=bxy=b^x is not well defined on the real numbers when b<0b<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)=2log⁡−2(−2)\log_{-2}(4) = 2 \log_{-2}(-2)

and it is easy to be temped to say that log⁡−2(−2)=1\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)=ln⁡(−2)ln⁡(−2)\log_{-2}(-2) = \dfrac{\ln(-2)}{\ln(-2)}

but now we can use complex numbers here.

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

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

which means that

ln⁡(−2)=ln⁡2+i(π+2πk)\ln (-2) =\ln 2 + i(\pi + 2\pi k)

thus we have

log⁡−2(−2)=ln⁡2+i(π+2πn)ln⁡2+i(π+2πm)\log_{-2}(-2) = \dfrac{\ln 2 + i(\pi + 2\pi n)}{\ln 2 + i(\pi + 2\pi m)}

for n,m∈Zn,m\in\mathbb{Z}.

So, setting n=mn=m gives us log⁡−2(−2)=1\log_{-2}(-2) = 1, but if n≠mn\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)\log_{-2}(-2) represents.

Amazing explanation !! Thank you :smile:

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