I'm genuinely happy to know that logarithms of negative numbers DO exist after all. The misconception about them not existing is far more enduring than that about square roots of negative numbers not existing. Neat video!
I don't know who you are or why you do this videos, but I thank you. My calculus course right now is not the most comfortable one so to say, and your guidance with these problems is much appreciated. Your explanations are thorough and unique, an aspect of your channel that I love to see. Thank you!! Keep it up!!
@@denn2501 yeah but you can say negative numbers arent real cos you cant have -2 apples. Imaginary numbers arent imaginary, I believe that it's a misnomer Edit: i missed the joke totally...
You are the best!! When i was in my undergrad advanced calculus, our teacher mentioned the fractional derivatives. I have read some things on the topic, but it would be awesome if u make a video on this topic because i dont really understand that much about that
Since a logarithm with negative number(s) has an infinite number of complex solutions, it stops being a function, let alone a nice smooth one, and turns into a mess, so it would make sense that negative logarithms aren't used, unless needed of as a curiosity. Maybe it could still be a function, if you always assumed m,n to be 0, or used some other method to pick one solution, though.
What breaks if we create a new op # that is to addition as addition is to multiplication so: a#b = ln(e^a+e^b) Then the “hashative” inverse of a is a +pi*i and the hashative identity H is a new number that approaches negative infinity. Is it screwed up by branches of ln on complex plane?
Well, the explanation was fine. But sorry, blackpenredpen, you should have stated that the other roots are extraneous, actually, hence one solution only.
So in getting the result, log₋₋₂(-8) = [3ln2 + (2m+1)πi]/[ln2 + (2n+1)πi] you used the complex values of both ln(-8) and ln(-2). Why didn't you use both when taking log₂(-8) and log₂(8)? Shouldn't those have (ln2 + 2nπi) in the denominator? In short, I have some doubts about whether your full final result is valid. The ultimate test is whether you can use that result as the exponent on the base of logarithms, and get the argument of those logarithms. Can you reassure us that that will work? Fred
I actually thought about that but I was just thinking the denominator is just ln2, which is a real number so we didn’t have to use that formula. But of course, any real number is also complex just like a=a+0i... so I don’t know anymore : )
@@blackpenredpen I will try to delve into this question a little deeper, when I have some time; seems to me at the moment, that it could go either way... Fred
Its possible to get negative input if the exponent is an odd number. However, if an exponent is an even number, such as 2, 4, 6, 8, etc., you will get an error. For natural logarithms, the input has to be positive.
okay so I have a question. if we have 4^x=-4 why cant we square both sides to get 4^2x=16 then but it under a logarithm log4^16=2x 4=2x x=2 please help
What's the real meaning of solving any equation? To get the value of x that satisfies a given expression? Or to get the value of x wich is included in the domain of a given function?
To see why we can't have log base (-2) of (-8) let's say it is equal to k. So (-2)^k = -8, BUT (-2)^(2*k/2) is also -8, because we can just multiply and divide by the same number. So in this example k = 3, but also k = 6/2. (-2) ^ (6/2) = sqrt [ (-2)^6 ] = sqrt [64] = 8 != -8. You see now, why we don't have negative bases for exponential functions? And from this fact and defenition of logarithm we get this restriction on logarithms also.
I think this is the same situation as [sqrt(-1)]^2 or e^[pi*sqrt(-1)] they're both undefined in the real world but have real values. You can get many examples just using sqrt(-1) instead of i in a complex expression that gives a real value. The problem is that not all the passages use real numbers so the solution doesn't exist in the real world but it's a complex number that's the equivalent of a real number in the complex numbers. So for example the value of log_-2(-8) is not the real number 3 but the complex number 3.
@blackpenredpen okay so I have a question. if we have 4^x=-4 why cant we square both sides to get 4^2x=16 then but it under a logarithm log4^16=2x 4=2x x=2 please help
the basic log properties allow you to just push your way through a forest of impossibility. log(-8) = 3 * log(-2), therefore if the base is -2, the result is 3 by the definition that log base x of x is 1, except of course for 0, as with most rules.
Tbf, the same reasoning allows us to write log(x) = log(x) + log(1) = log(x) + 2 * m * Pi * i, where x > 0. Well it does not mean 2 * m * Pi * i = 0 though since it just shows that the log isn’t properly defined 🤔
No, an imaginary term can still be present over positive real numbers, but in that case it's even multipliers of pi instead of odd ones. The real solution is just the one at m=0.
I am studying complex analysis at university--oh boy do I not get troubled understanding Laplacian and the Cauchy-Riemannnnnnnnnnnnnn equations. If I could be helped:
Pranav Suren By the standard log function, I assume you mean the natural logarithm? If so, then the natural logarithm is undefined for negative numbers, but the logarithm for negative bases IS well defined, but for real-valued codomains, its domain is not the entire set of real numbers. This is what is meant. It is not defined for every input either, but that is irrelevant. The video is giving an example of a logarithmic operator which is well-defined for negative inputs AND is real-valued. In this, BPRP is not wrong. He never claimed the natural logarithm is defined for negative numbers. In fact, the sentence "it is undefined for negative numbers" is nonsensical because it is not mathematically coherent to talk about a function being undefined without first specifying some domain and codomain.
Angel Mendez By that u meant that the logarithmic functions with neg basis are not continious functions They can't be defined for some real numbers . But the example bprp gave was not defined for all real numbers as well ,
I don't understand why in your video about the sum from n=1 to infinity of sin(x)/x why you only included one answer for the sum whem there are infinite?
My solution to 2^k = -8: 8*2^(k-3) = 8*e^((k-3)*ln2) if (k-3)*ln2=i*pi, then 8*e^(i*pi) = -8. k=i*pi/ln2+3 There. I started before he got to complex numbers.
Really interesting like always, a big thanks to you! Before looking to your video, I already tried this, I did it quit different: Let A = Log₂(-8) = (ln -8 )/(ln 2) = [(ln 8 + ln -1)]/(ln 2) but with Euleur’s formula, e^iπ=-1 ; A = (ln8 + iπ)/(ln 2). I’m only a 11th grader, I haven’t even see that in class yet but I hope that’s not so false;) (sorry for my bad English)!
Can we really say that log exists if it has multiple answers? In contrast for limit, we say the limit of a function does not exist if it gives a different value from the "right" or "left"
Kinda random, but what if you take the limx->0 ln(x) fang you make this -infinity? Since it would be e^k=0, neg exponent mean reciprocal, so it could be 1/e^(-infinity)? I’m not in calc yet so I’m not sure if there’s a rule to this limit or anything, but just curious
A good reply is that although the value exists, the function is not continuous at the point, which makes it invalid, which makes the vid's argument (complex pun not intended) valid (:) double smile
if the basis is positive, is clearly that log can not be real...but, you are correct, if the basis is negative, depending of the argument, the expoent can be real!!!!! All depends how you define the exponential logarithmic function
log-2(-8) doesn't make any sense because base of the exponentional function should be always positive otherwise there is a problem (-2)^3 = -8 (-2)^3 = (-2)^6/2 = 64^1/2 = 8 8 = -8
Can derivatives of functions exist at endpoints? I argue, with support from online sources that derivatives cannot exist at endpoints because the limit does not exist from both sides because the function does not exist from both sides. However, my math teacher asserts that derivatives must be evaluated within the domain of a function so one-sides derivatives become the actual derivative. The question is whether or not f’(x) exists at the endpoint of f(x). This is in the context of AP calc BC. Example: what is dy/dx of x^1.5+y^1.5=10 at x=0
I was thinking some years ago : " If the ln of a negative number doesn't exist , what does it mean to take the primitive function of y= 1/x on the negatives . The graph doesn't exist ?
Hi I am a math teacher in University of Garmian in Kalar a small part of the Kurdistan Region-Iraq. And I have a one problem [ Let a and b be two real number such that a
![Blox Fruits Dragon Rework Update [Full Stream]](http://i.ytimg.com/vi/EqDAp8udhm0/mqdefault.jpg)
![Logarithms of Negative BASES and Negative Arguments! [ Log of Complex Numbers z ]](/img/1.gif)
Thank you once again!
On here you are !!!
You’re welcome!!!
* * * *
* * * *
@@blackpenredpen Sir please explain how you got (2m+1)πi ?
I haven't study complex numbers completely just know the basics.
@@samajlo4336 see the link in the description, there is a link to the explanation
Doctor: Log (neg) isnt real, it cant hurt you.
Log (neg): log-2(-8)=3
Breno Sanches Hahahhahaha!!!
I think the burn hurt quite a bit.
@@blackpenredpen iota will not help here ???????
Imaginary solutions for negative logarithms ????
@@blackpenredpen try iota as a base for logarithms by Euler identity by taking log on both side
Don't worry, it's just imaginary xD
I'm genuinely happy to know that logarithms of negative numbers DO exist after all. The misconception about them not existing is far more enduring than that about square roots of negative numbers not existing. Neat video!
I don't know who you are or why you do this videos, but I thank you. My calculus course right now is not the most comfortable one so to say, and your guidance with these problems is much appreciated. Your explanations are thorough and unique, an aspect of your channel that I love to see. Thank you!! Keep it up!!
Jose Gomez you’re very welcome. Thank you for the nice comment!
I think he’s a college math teacher. Some of his videos reference “problem whichever from the homework”
They said log(neg) can't be real
Imaginary Numbers: "Allow us to introduce ourselves."
X Ry
Wait
but imaginary number is not "real" number so....
"is i a joke to you?"
@@denn2501 yeah but you can say negative numbers arent real cos you cant have -2 apples. Imaginary numbers arent imaginary, I believe that it's a misnomer
Edit: i missed the joke totally...
Get real
Something: No solution
Mathematician: But imagine if it did lol
Wicked people: (-2) base log (-8) is not possible.
Imaginary Number: Hold my beer.
hahaha
it's 3
ln(-8)=ln[(-2)^3]=3ln(-2) then log-2(-8)=3
The relief I felt when that last equation equaled 3 was euphoric.
Someone tells me you can’t take the log of a negative. Me: “not reeeaalllly...”
not *real* -ly...
You are the best!! When i was in my undergrad advanced calculus, our teacher mentioned the fractional derivatives. I have read some things on the topic, but it would be awesome if u make a video on this topic because i dont really understand that much about that
High School: Log of a negative number can't be real
College:
Me: Oh God oh fr*ck I'm literally shaking
kkkkkkkkk
Damn, you censored fr*ck
Lel I'm in 10th grade and can do logarithms (positive bases) of negative numbers and of complex numbers.
@@pedromendz double censorship
Dr paymn saying woah was lit😂
If this channel has taught me anything, it's that finding no solutions just means you haven't checked the complex numbers yet.
All those dislikes from those Math Teachers.
Lovely explanation, I don't get why are you
so underrated , btw thank you so much you just nailed it, man ......
You are awesome man, thanks for all your work.
Greetings from a Maths student of Spain !
Donde estudias?
Since a logarithm with negative number(s) has an infinite number of complex solutions, it stops being a function, let alone a nice smooth one, and turns into a mess, so it would make sense that negative logarithms aren't used, unless needed of as a curiosity. Maybe it could still be a function, if you always assumed m,n to be 0, or used some other method to pick one solution, though.
Thankyou brother, you literally just cleared all the restlessness!!!!
thanx...
Euler's Identity is one of history's greatest mathematical discoveries.
knowing imaginary numbers is like knowing another dimension.
0:03 Dr. Peyam reminds me of Jebediah Kerman
at 9:25, just let m=n simplify (2m+1)pi*i and you get ln(8)/ln(2) which is equal to 3
In the imaginary world, things get real that couldn't be imagined in the real world.
What breaks if we create a new op # that is to addition as addition is to multiplication so:
a#b = ln(e^a+e^b)
Then the “hashative” inverse of a is a +pi*i and the hashative identity H is a new number that approaches negative infinity. Is it screwed up by branches of ln on complex plane?
e^(pi)i=-1
//ln both sides
(pi)i=ln(-1)
When you don't know what to put in the "#"
*#complexlogarithm*
*#MathForFun*
*#dearsubscriber*
#mathetinfo 😉😁
care need to be taken when dealing with complex logarithmic.
8:17 only n=0 are the correct answer to (-2)^k=-8, other values of n are extraneous root
Well, the explanation was fine.
But sorry, blackpenredpen, you should have stated that the other roots are extraneous, actually, hence one solution only.
So in getting the result,
log₋₋₂(-8) = [3ln2 + (2m+1)πi]/[ln2 + (2n+1)πi]
you used the complex values of both ln(-8) and ln(-2).
Why didn't you use both when taking log₂(-8) and log₂(8)? Shouldn't those have (ln2 + 2nπi) in the denominator?
In short, I have some doubts about whether your full final result is valid.
The ultimate test is whether you can use that result as the exponent on the base of logarithms, and get the argument of those logarithms.
Can you reassure us that that will work?
Fred
I actually thought about that but I was just thinking the denominator is just ln2, which is a real number so we didn’t have to use that formula. But of course, any real number is also complex just like a=a+0i... so I don’t know anymore : )
@@blackpenredpen I will try to delve into this question a little deeper, when I have some time; seems to me at the moment, that it could go either way...
Fred
@@ffggddss did you figure it out?
Real world problems require complex solutions.
Only if exponent is odd and not even
But doesn't using infinite soultions make log not a function but a relation instead. We restricted the range of arcsin, can't we do it for log too?
Surely you can. By using primary solution with m=0. That function is written as Ln(x) instead of ln(x) (yeah just using the capital letter)
Maybe make a quickie on Log (base-i) of 2 or Log (base-Z) of x where Z complex and x is real, just to complete the picture!
John Napier would be happy now :) LOL
Mandhan Academy yup!!
Wow that's so cool! Thank you!
Its possible to get negative input if the exponent is an odd number.
However, if an exponent is an even number, such as 2, 4, 6, 8, etc., you will get an error.
For natural logarithms, the input has to be positive.
okay so I have a question.
if we have 4^x=-4
why cant we square both sides to get 4^2x=16
then but it under a logarithm log4^16=2x
4=2x
x=2
please help
What's the real meaning of solving any equation?
To get the value of x that satisfies a given expression?
Or to get the value of x wich is included in the domain of a given function?
and probability bro can you please make a video. .. your explanations are the best 😉👍
I feel smarter just by listening this man speak
To see why we can't have log base (-2) of (-8) let's say it is equal to k. So (-2)^k = -8, BUT (-2)^(2*k/2) is also -8, because we can just multiply and divide by the same number. So in this example k = 3, but also k = 6/2.
(-2) ^ (6/2) = sqrt [ (-2)^6 ] = sqrt [64] = 8 != -8. You see now, why we don't have negative bases for exponential functions? And from this fact and defenition of logarithm we get this restriction on logarithms also.
I think this is the same situation as [sqrt(-1)]^2 or e^[pi*sqrt(-1)] they're both undefined in the real world but have real values. You can get many examples just using sqrt(-1) instead of i in a complex expression that gives a real value. The problem is that not all the passages use real numbers so the solution doesn't exist in the real world but it's a complex number that's the equivalent of a real number in the complex numbers. So for example the value of log_-2(-8) is not the real number 3 but the complex number 3.
Calculator showing error😂
I don’t understand this at all but I love hearing you talk about it
Dr. Peyam 😂 wowww.....😆😆😆😆
In 3:07 you said logarithm of zero is not defined and I also read somewhere that it isn't defined even in complex domain. Why is that?
It's 1/infinity
how are you going to exponentiate a number to 0?
The CHAIN RULE!!!! That blasted Chen Lu thing had me stump for ages. Only now reading some of the comments did I understand it :D
Log is just a tool, it can be used as it fits, despite the typical existence conditions.
This comment is slightly underrated
0:04 when the bullies are on your case
@blackpenredpen
okay so I have a question.
if we have 4^x=-4
why cant we square both sides to get 4^2x=16
then but it under a logarithm log4^16=2x
4=2x
x=2
please help
You can't square a negative number in an equation . You can square only if the number is positive
This is unreal!!!
I have a question : why ln(2) in the denominator is not also multivalue function ?
the basic log properties allow you to just push your way through a forest of impossibility. log(-8) = 3 * log(-2), therefore if the base is -2, the result is 3 by the definition that log base x of x is 1, except of course for 0, as with most rules.
I am a student of class 12,Bangladesh. Thanks man! I was confused for this question.you have made it easy...
1:47 1 isn't it after an isn't it wow next lvl
Tbf, the same reasoning allows us to write
log(x) = log(x) + log(1) = log(x) + 2 * m * Pi * i, where x > 0.
Well it does not mean 2 * m * Pi * i = 0 though since it just shows that the log isn’t properly defined 🤔
The ipi is from the argument being negative though
No, an imaginary term can still be present over positive real numbers, but in that case it's even multipliers of pi instead of odd ones. The real solution is just the one at m=0.
Yeah, everything can be expanded to complex world, because ln4 = 2ln(-2) ...
The last part was amazing
I am studying complex analysis at university--oh boy do I not get troubled understanding Laplacian and the Cauchy-Riemannnnnnnnnnnnnn equations. If I could be helped:
Need help?
I just finished complex analysis
@@marks9618 can u help me with my maths GCSE
Skipsotz Gaming sure!
@@marks9618 how can you help me, do you have discord or something
9:45
“[...] they [non-math savvy people] want to stay in the real world”
Sheesh, savage !
Guess we are math savvy.
1:47 *in Patrick Bateman's voice*
YES IT IS!
Trop Fort ! Un vrai pédagogue ! U are the Best
Mind = Blown
Me at beginning of video: Doesn’t all negative logs have a complex solution as ln(-1) = i*pi
Me at end:
Alrighty then.
Who said log(neg) can't be real ??
Yeah that's pretty much what this video about though .
Pranav Suren By the standard log function, I assume you mean the natural logarithm? If so, then the natural logarithm is undefined for negative numbers, but the logarithm for negative bases IS well defined, but for real-valued codomains, its domain is not the entire set of real numbers. This is what is meant. It is not defined for every input either, but that is irrelevant. The video is giving an example of a logarithmic operator which is well-defined for negative inputs AND is real-valued. In this, BPRP is not wrong. He never claimed the natural logarithm is defined for negative numbers. In fact, the sentence "it is undefined for negative numbers" is nonsensical because it is not mathematically coherent to talk about a function being undefined without first specifying some domain and codomain.
Angel Mendez
By that u meant that the logarithmic functions with neg basis are not continious functions
They can't be defined for some real numbers .
But the example bprp gave was not defined for all real numbers as well ,
@@angelmendez-rivera351 actually no one mentioned natural logarithm.
The people who say “Chen Lu” as “Chain Rule”
Thanks man 😊
Got my all doubt clear
It's amazing, I like your youtube channel !!!!
Can you make a video of fractional calculus
Are log sub b of positive number have complex solution?
I don't understand why in your video about the sum from n=1 to infinity of sin(x)/x why you only included one answer for the sum whem there are infinite?
My solution to 2^k = -8:
8*2^(k-3) = 8*e^((k-3)*ln2)
if (k-3)*ln2=i*pi, then 8*e^(i*pi) = -8.
k=i*pi/ln2+3
There.
I started before he got to complex numbers.
Really interesting like always, a big thanks to you! Before looking to your video, I already tried this, I did it quit different:
Let A = Log₂(-8) = (ln -8 )/(ln 2) = [(ln 8 + ln -1)]/(ln 2) but with Euleur’s formula, e^iπ=-1 ; A = (ln8 + iπ)/(ln 2).
I’m only a 11th grader, I haven’t even see that in class yet but I hope that’s not so false;) (sorry for my bad English)!
e^(2k+1)iπ = -1, k is any integer
I ♥ complex world
There is nothing to stop your head
Atlast someone who explains the reason 🤩🤩
I love that
sqrt(x)=-1 does not exist
sqrt(x)=-2 does exist
I prefer writing odd numbers as 2n-1 because this way you can get all the positive odd numbers for natural ns
Can we really say that log exists if it has multiple answers? In contrast for limit, we say the limit of a function does not exist if it gives a different value from the "right" or "left"
Kinda random, but what if you take the limx->0 ln(x) fang you make this -infinity? Since it would be e^k=0, neg exponent mean reciprocal, so it could be 1/e^(-infinity)?
I’m not in calc yet so I’m not sure if there’s a rule to this limit or anything, but just curious
Helped me a lot!🖤Thanks ❤May Allah help you and Guide you and best wishes from Bangladesh 🇧🇩✌
A good reply is that although the value exists, the function is not continuous at the point, which makes it invalid, which makes the vid's argument (complex pun not intended) valid
(:) double smile
You cant say X>0 in complexe wish ln (-x) when x>0 dosnt exist in C because x= a+ib.
Its not general in C
is that a greenpen I see on this blackpenredpen channel!?!?
log(-2,-8) = ln(-8)/ln(-2) = 1.09284065 - 0.420787248 i (assuming google can do complex arithmetic).
You said log(-2,-8) = 3.
These are different.
if the basis is positive, is clearly that log can not be real...but, you are correct, if the basis is negative, depending of the argument, the expoent can be real!!!!! All depends how you define the exponential logarithmic function
sir could you please explain why z^n = conjugate of z , has n+2 solutions with z as a complex number?
Bprp can you make a video on log 0
I found (for second problem) that k=3 in all cases when m = 3n+1, not only when m=1 & n=0
If my tutors were like u😣😣
i cannot understand the end part. For what numbers of m and n, the final result can be 3 ???
He answered the question. For m=1 and n=0, the final answer can be 3.
Is it really necessary for the complex logarithm to be strictly for base e or nah?
So if you take log base 2i of a negative number in the complex plane, do you always get a real answer?
log-2(-8) doesn't make any sense because base of the exponentional function should be always positive
otherwise there is a problem
(-2)^3 = -8
(-2)^3 = (-2)^6/2 = 64^1/2 = 8
8 = -8
Same way eigenvectors that are imaginary indicate spin in a higher dimension
Log(-2)-8=Log(2i^2)8i^6= Log(2i^2)(2i^2)^3=3
Is this set of solutions actually dense in the complex plane?
Can derivatives of functions exist at endpoints? I argue, with support from online sources that derivatives cannot exist at endpoints because the limit does not exist from both sides because the function does not exist from both sides. However, my math teacher asserts that derivatives must be evaluated within the domain of a function so one-sides derivatives become the actual derivative. The question is whether or not f’(x) exists at the endpoint of f(x). This is in the context of AP calc BC.
Example: what is dy/dx of x^1.5+y^1.5=10 at x=0
I was thinking some years ago : " If the ln of a negative number doesn't exist , what does it mean to take the primitive function of y= 1/x on the negatives . The graph doesn't exist ?
Hi I am a math teacher in University of Garmian in Kalar a small part of the Kurdistan Region-Iraq. And I have a one problem [ Let a and b be two real number such that a
Do you use Kurdish letters as variables in mathematics?
Like your videos 👍
Sir function of log having base and value are are not defined??
can you proof that rule?
Dark see description
@@blackpenredpen thank you!