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Why can't my calculator do log -2(4) ?

I don't get why.

Original post by lhh2003

I don't get why.

Because logarithms cannot take in a negative number. (At A-Level, anyway)

Reply 2

Gregorius

Original post by lhh2003

I don't get why.

Needs a bigger calculator: https://www.wolframalpha.com/input/?i=Log%5B-2%2C+4%5D

Reply 3

Physicsqueen

Original post by RDKGames

Because logarithms cannot take in a negative number. (At A-Level, anyway)

Wait Logs can take negative values? (I’m not questioning the statement, just curious as I’ve not come across it, I’m at A-level), how does it work? Is it to do with complex numbers?

Reply 4

lhh2003

Original post by Gregorius

Needs a bigger calculator: https://www.wolframalpha.com/input/?i=Log%5B-2%2C+4%5D

(-2)^2 = 4 , so what is with all the imaginary numbers in that calculator ? I don't do FM btw.

Reply 5

username3331778

Original post by Physicsqueen

Wait Logs can take negative values? (I’m not questioning the statement, just curious as I’ve not come across it, I’m at A-level), how does it work? Is it to do with complex numbers?

Write the number in complex exponent form and it should be obvious what to do next

Reply 6

RDKGames

Study Forum Helper

Original post by Physicsqueen

Wait Logs can take negative values? (I’m not questioning the statement, just curious as I’ve not come across it, I’m at A-level), how does it work? Is it to do with complex numbers?

Yes, it's to do with complex numbers.

Any complex number can be written as

z = |z|e^{i \arg z}

therefore, taking logs means

\log z = \log(|z|e^{i \arg z}) = \log |z| + \log (e^{i \arg z}) = \log |z| + i \arg z

But since

\arg z

can take on multiple values (we call them branches) then we can agree on the so-called principal value of

\log z

. This occurs when we only take the argument of

z

to be over the interval

-\pi < \arg z \leq \pi

Hence, we have the natural log of a complex number

z

along the principal branch as:

\mathrm{Log} \ z = \log |z| + i \mathrm{Arg} \ z

where the capital L and A letters symbolise that we are taking the principal branch value.

Anyway, once we have this established, it is clear that for

z = -2

we have

\mathrm{Log} \ (-2) = \log 2 + \pi i

If you want other branches, just add/subtract

2\pi

from the argument; hence really, it means that

\log (-2)

takes on infinitely many values of the form

\log (-2) = \log 2 + \left( \pi + 2 \pi n \right) i

where

n \in \mathbb{Z}

(edited 3 years ago)

Reply 7

Gregorius

Original post by lhh2003

(-2)^2 = 4 , so what is with all the imaginary numbers in that calculator ? I don't do FM btw.

Life gets more complicated when you take ordinary every day functions and start to think of them in terms of complex numbers (which is something that you have to do as soon as you start thinking about logarithms involving negative numbers).

Remember that answer that Wolfram gave:

https://www.wolframalpha.com/input/?i=Log%5B-2%2C+4%5D

Feed it back, raising (-2) to its power:

https://www.wolframalpha.com/input/?i=%28-2%29%5E%28Log%5B4%5D%2F%28I*Pi+%2B+Log%5B2%5D%29%29

and you get 4. Sort of suggest that there is more than one answer to the original question.

It turns out that you have to be very careful about what these sorts of things mean. For example, you can write a number like -2 as 2 x exp(-i x Pi), and therefore log(-2) = log(2) - i x Pi (where we're using natural logarithms). If you recall the formula for changing between bases of logarithms, you should now see where the imaginary numbers that Wolfram gave come from!

But why doesn't Wolfram give the "obvious" answer of 2 to the question "what is log(-2, 4)"? It's because complex logarithms are "many valued" unless you start doing things to make them single valued. When you start doing these things (in a particular way), all of a sudden the single values that you get are not necessarily the ones you might think you should get.

Reply 8

B_9710

Original post by RDKGames

Yes, it's to do with complex numbers.

Any complex number can be written as

z = |z|e^{i \arg z}

therefore, taking logs means

\log z = \log(|z|e^{i \arg z}) = \log |z| + \log (e^{i \arg z}) = \log |z| + i \arg z

But since

\arg z

can take on multiple values (we call them branches) then we can agree on the so-called principal value of

\log z

. This occurs when we only take the argument of

z

to be over the interval

-\pi < \arg z \leq \pi

Hence, we have the natural log of a complex number

z

along the principal branch as:

\mathrm{Log} \ z = \log |z| + i \mathrm{Arg} \ z

where the capital L and A letters symbolise that we are taking the principal branch value.

Anyway, once we have this established, it is clear that for

z = -2

we have

\mathrm{Log} \ (-2) = 2 - \dfrac{\pi}{2}i

If you want other branches, just add/subtract

2\pi

from the argument; hence really, it means that

\log (-2)

takes on infinitely many values of the form

\log (-2) = 2 - \left( \dfrac{\pi}{2} + 2 \pi n \right) i

where

n \in \mathbb{Z}

You made a mistake for log(-2)

Reply 9

RDKGames

Study Forum Helper

Original post by B_9710

You made a mistake for log(-2)

Whoops!

Reply 10

B_9710

Original post by RDKGames

Whoops!

Even still

Reply 11

Physicsqueen

Original post by RDKGames

Yes, it's to do with complex numbers.

Any complex number can be written as

z = |z|e^{i \arg z}

therefore, taking logs means

\log z = \log(|z|e^{i \arg z}) = \log |z| + \log (e^{i \arg z}) = \log |z| + i \arg z

But since

\arg z

can take on multiple values (we call them branches) then we can agree on the so-called principal value of

\log z

. This occurs when we only take the argument of

z

to be over the interval

-\pi < \arg z \leq \pi

Hence, we have the natural log of a complex number

z

along the principal branch as:

\mathrm{Log} \ z = \log |z| + i \mathrm{Arg} \ z

where the capital L and A letters symbolise that we are taking the principal branch value.

Anyway, once we have this established, it is clear that for