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

Watch
lhh2003
Badges: 16
Rep:
?
#1
Report Thread starter 4 months ago
#1
I don't get why.
0
reply
RDKGames
Badges: 20
Rep:
?
#2
Report 4 months ago
#2
(Original post by lhh2003)
I don't get why.
Because logarithms cannot take in a negative number. (At A-Level, anyway)
0
reply
Gregorius
Badges: 14
Rep:
?
#3
Report 4 months ago
#3
(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
1
reply
Physicsqueen
Badges: 17
Rep:
?
#4
Report 4 months ago
#4
(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?
0
reply
lhh2003
Badges: 16
Rep:
?
#5
Report Thread starter 4 months ago
#5
(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.
0
reply
username3331778
Badges: 16
Rep:
?
#6
Report 4 months ago
#6
(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
0
reply
RDKGames
Badges: 20
Rep:
?
#7
Report 4 months ago
#7
(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}.
Last edited by RDKGames; 4 months ago
1
reply
Gregorius
Badges: 14
Rep:
?
#8
Report 4 months ago
#8
(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/?...g%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.
0
reply
B_9710
Badges: 17
Rep:
?
#9
Report 4 months ago
#9
(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)
0
reply
RDKGames
Badges: 20
Rep:
?
#10
Report 4 months ago
#10
(Original post by B_9710)
You made a mistake for log(-2)
Whoops!
0
reply
B_9710
Badges: 17
Rep:
?
#11
Report 4 months ago
#11
(Original post by RDKGames)
Whoops!
Even still
0
reply
Physicsqueen
Badges: 17
Rep:
?
#12
Report 4 months ago
#12
(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) = \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}.
Thank you
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (49)
16.23%
I'm not sure (8)
2.65%
No, I'm going to stick it out for now (103)
34.11%
I have already dropped out (4)
1.32%
I'm not a current university student (138)
45.7%

Watched Threads

View All