You use the complex logaritm. It's a fonction only if the antecedent is in [0,2.pi[ (main determination). Here by setting -5 = (5,pi) (polar), X=a+i.b and 5=(5,2k.pi) with k € Z, the solutions are : a = ( (ln(5))^^2 + 2.k.(pi)^^2 ) / ( (ln(5))^^2 + (pi)^^2 ) and b = 2k.pi/ln(5) + pi/ln(5) (ln(5)^^2 + 2k.(pi)^^2)/(ln(5)^^2 +(pi)^^2). For k=0 the result is ln(5) / ln(5)+i.pi.
Well, let's just follow the rules. (-5)^x = 5 ln((-5)^x) = ln(5) x*ln(-5) = ln(5) x = ln(5) / ln(-5) = 0.208 - i*0.406 That checks as correct. No tricks required - just do the arithmetic.
@@Nikioko if you write only the answer, x = ln(5) / ln(-5). it is also write as you don't have calculator and it is a short question and x = ln(5) / ln(−5) = ln(5) / (ln(5) + πi). it is just more explanation and at the end, you get same result. As 'ln' function has only domain of positive real number and the domain of negative real number means complex number. ln(−5) = ln(5) + ln(−1)= ln(5) + πi , Here you also don't explain how do you get the result pi * i . you need elaborate explanation as well.
Gracias, 2 métodos q un gran profesor nos enseñó, hace varios años pero con ecuaciones trigonométricas y números trascendentes... Pero está excelente su explicación, y el Buen recuerdo de ese gran profesor de matemáticas, se renovó.
This provides just one solution of the infinite solutions that this equation has. All solutiona are included in: x=ln 5 / (ln 5 +i-pi.(2n + 1)), where n belongs to Z . Anyway, the solution is easiily found by using polar coordinates as already posted.
You're twice wrong. Log[n](x) is DEFINED on the field of real numbers |R only for x > 0 & 1 ≠ n > 0. Whereas it has a generalization on dual & complex numbers & can be defined to have generalization on many more other sets. The question means just that. Or otherwise there's no solution (on reals), which is never a complete or meaningful answer in mathematics. He did the most basic & adequate thing by using that quality. I guess you'd be surprised if told square root of -1 exists & is equal to i.
@@stardustwight1895 I am not wrong. You need to check the solution to truly verify that indeed it works. Replacing X by the solution must equate 5, which it does not. Hence, no solution. A meaningful answer must be proved, so prove that by replacing x with the solution produces a 5. If it does not fit then it is not a solution and it is useless
That's not the only solution. If you use Euler's identity, use it properly. -1 = e^(i Pi) is only one special case. Generally, -1 = e^(i (2k + 1)Pi) for k in Z. That gives you countable infinity of solutions in complex numbers.
Xlog(-5)=log5: x=log5/log(-5): but -ve of log is undefined, as right side is +ve, left side should be +ve. If x is even no. And (-5)^x=5, it will true. So x=2×0.5=1 or 😅(5i)^x=5 : x=1/((logi/log5)+1)
@@saketashol6728no, not that either. They ask you, if you make the interview, what you see yourself doing there. What departments or professors drew your interest. How you will be of service to the community. Sometimes they WILL ask you about your particular work if you have anything in your background and file that is unusual. Like, you helped develop a program for treating cholera or your local church created a system to distribute clothing to the needy. Or you worked in a law office and studied some preliminary work. That’s what most of those interviews are like. Unless you’re a performing arts person and it’s graduate. That’s an audition.
@@Nikioko (-5)ⁿ=5 [(-1)×(5)]ⁿ=5 (-1)ⁿ×(5)ⁿ=5 (-1)ⁿ=5/5ⁿ (-1)ⁿ=5¹-ⁿ (-1)ⁿ×(-1)ⁿ=(5¹-ⁿ)×(-1)ⁿ [(-1)ⁿ]²=(5¹-ⁿ)×(-1)ⁿ; but (-1)ⁿ=5¹-ⁿ (-1)²ⁿ=(5¹-ⁿ)×(5¹-ⁿ) [(-1)ⁿ]²=(5¹-ⁿ)² [(-1)²]ⁿ=(5²)¹-ⁿ [1]ⁿ=(25)¹-ⁿ 1ⁿ=25¹-ⁿ but 1ⁿ=1 for every n. Therefore, 25¹-ⁿ=1 25/25ⁿ=1 25=25ⁿ 5²=(5²)ⁿ 5²=5²ⁿ 2=2n n=1 is a solution but, (-5)¹=-5 and -5≠5 then n=1 is a absurd. Therefore, do not exist a solution. ////////////////////////////////////////////////// retake: (-1)ⁿ=5¹-ⁿ if exist a n such that 1ⁿ≠1, so: a=a (assuming this is true) and a≠0 a/a=1 (a/a)ⁿ=1ⁿ (a/a)ⁿ≠1 aⁿ/aⁿ≠1 aⁿ≠aⁿ a≠a (a absurd conclusion) when, a=0 0=0 0ⁿ=0ⁿ is true for every n≠0 when, n=0, n is not a incognite and 0ⁿ=0⁰ is indefinite. Therefore, 1ⁿ≠1 is impossible. (remember that is -1≠1) ////////////////////////////////////////////////// Therefore, I can't see (-5)ⁿ=5 with a possible solution.
That's relevant here only if we consider -5^x to be a multi-valued function. But i*2*pi*n plays a role in another part of the solution, which the author of the video missed: exp(x * Log(-5)) = exp(Log(5)) x * Log(-5) = Log(5) + i*2*pi*n x = (Log(5) + i*2*pi*n) / Log(-5). And if we consider -5^x to be a multi-valued function, then the solution is: exp(x * (Log(-5) + i*2*pi*m)) = exp(Log(5)) x * (Log(-5) + i*2*pi*m) = Log(5) + i*2*pi*n x = (Log(5) + i*2*pi*n) / (Log(-5) + i*2*pi*m). Log here represents the principal branch of the logarithm, i.e. Log(-5) = Log(5) + i * Pi.
because log with base 10 is commonly used worldwide. when we write "log 2" this expression directly means that the base is 10. you can say we use it conventionally.
@@syther836 using log base 5 would significantly simplify the expression immediately, so using log 10 because it is used commonly worldwide is a lame excuse. Why not to add sin and cos there just because it is used worldwide?
You should have explain that first you must know that i^2 is equal to -1 because i=√(-1) and √(-1)* √(-1) cancels out the √ and therefore gives you a -1.
Solving one of those equations that would never occur in the real world. I could, of course, learn how to do that and than keep and keep practicing it over my lifetime, just in case there's this 1 in a billion chance I will ever need it. Or I could learn it once, when I actually need to in one of my next 10 lives for half an hour, to forget it then again thereafter because of how useless it is. I'm all for math in applied settings, but endlessly solving pointlessly constructed, completely unrealistic equations, with methods that you'll have to practice all the time because you'll have forgotten half of it in 2 years, because that's how useful it actually is in life, just seems like playing sudoku as a hobby, rather than actually using math for something cool and real world. Math practice that actually makes sense because you need it all the time, is to practice how to quickly calculate all kinds of things in your head, developing shortcuts to quick approximations and these kinds of things. Solving unrealistic equations is just such a useless 'school skill' that it actually annoys me. It's like a hobby, like playing sudoku, it doesn't make you smart and it doesn't give you any real useful skills. It just wastes your time. And if you stop practicing it for 2 years, you'll have forgotten half of it anyway, because you just never need it.
crap! (-5)^a can be defined only for a = p/q (as f:R-->R) (-5)^(sqrt(2)) = ?? ln(-1) does not exist! or == 0 2ln(-1) = ln(-1)^2 = ln 1 = 0 so ln-1 = 0 (if exists!) so 0 = ln-1 = ln(e^(i*pi*x) = i*pi *x only for x=0 and this is not a solution!
@zenekk9684 He went over that. There can be no real solution, so we turn to Euler's identity. Is there some reason not to? Was it used wrongly? Is there also no solution using log base 10? I read your post three times and still don't know what you mean. To critique a demonstration, go step by step and make sure your notation is clear, just like in the video.
So basically there is a solution group because we need to make it n* i* pi whereas the solutions makes no sense at all we put a rotation indicator in devisor . Nonetheless the analysis might be helping in a more complicated and realistic situation.
You use the complex logaritm. It's a fonction only if the antecedent is in [0,2.pi[ (main determination).
Here by setting -5 = (5,pi) (polar), X=a+i.b and 5=(5,2k.pi) with k € Z, the solutions are :
a = ( (ln(5))^^2 + 2.k.(pi)^^2 ) / ( (ln(5))^^2 + (pi)^^2 ) and
b = 2k.pi/ln(5) + pi/ln(5) (ln(5)^^2 + 2k.(pi)^^2)/(ln(5)^^2 +(pi)^^2).
For k=0 the result is ln(5) / ln(5)+i.pi.
bravo!
Can you post a video? Or perhaps a post in some forum? I don't quite understand the flattened symbols here. I mean a pic.
What symbol do you not understand?
Often I write Ln for the complex logarithm and ln for real logarithm (normal).
@@andretewem3385 "^^", what does it mean? 😅
@@andretewem3385 "^^" power of 2?
@0:5:33 you may not use log in complex domain in the way you did it here.
Well, let's just follow the rules.
(-5)^x = 5
ln((-5)^x) = ln(5)
x*ln(-5) = ln(5)
x = ln(5) / ln(-5) = 0.208 - i*0.406
That checks as correct. No tricks required - just do the arithmetic.
You forgot some stuff:
ln(−5) = ln(5) + ln(−1)= ln(5) + πi
Therefore,
x = ln(5) / ln(−5) = ln(5) / (ln(5) + πi).
@@Nikioko Ah, good point. Thanks.
@@Nikioko if you write only the answer, x = ln(5) / ln(-5). it is also write as you don't have calculator and it is a short question and x = ln(5) / ln(−5) = ln(5) / (ln(5) + πi). it is just more explanation and at the end, you get same result. As 'ln' function has only domain of positive real number and the domain of negative real number means complex number. ln(−5) = ln(5) + ln(−1)= ln(5) + πi , Here you also don't explain how do you get the result pi * i . you need elaborate explanation as well.
Gracias, 2 métodos q un gran profesor nos enseñó, hace varios años pero con ecuaciones trigonométricas y números trascendentes... Pero está excelente su explicación, y el Buen recuerdo de ese gran profesor de matemáticas, se renovó.
This provides just one solution of the infinite solutions that this equation has. All solutiona are included in: x=ln 5 / (ln 5 +i-pi.(2n + 1)), where n belongs to Z . Anyway, the solution is easiily found by using polar coordinates as already posted.
Just use De Moivre's Theorem. Set -5 = e^(ln 5 + pi*i). It is immediately obvious then x = ln 5 / (ln 5 + pi*i).
What's happen to the rules, you keep violating them.. log(n) valid if n >=1, otherwise your analysis and solutions are worthless and invalid
You're twice wrong. Log[n](x) is DEFINED on the field of real numbers |R only for x > 0 & 1 ≠ n > 0. Whereas it has a generalization on dual & complex numbers & can be defined to have generalization on many more other sets.
The question means just that. Or otherwise there's no solution (on reals), which is never a complete or meaningful answer in mathematics.
He did the most basic & adequate thing by using that quality.
I guess you'd be surprised if told square root of -1 exists & is equal to i.
@@stardustwight1895 I am not wrong. You need to check the solution to truly verify that indeed it works. Replacing X by the solution must equate 5, which it does not. Hence, no solution. A meaningful answer must be proved, so prove that by replacing x with the solution produces a 5. If it does not fit then it is not a solution and it is useless
That's not the only solution. If you use Euler's identity, use it properly. -1 = e^(i Pi) is only one special case. Generally, -1 = e^(i (2k + 1)Pi) for k in Z. That gives you countable infinity of solutions in complex numbers.
I recommend,-1, sub set, thanks for the math lesson, try hard at thinking, encouragement, great challenge test
Thank you for taking the time to make this video. Much appreciated. ❤
You are so welcome! Glad it was helpful. 🙏🙏🤩🤩
That was great Man! Two methods; very well explained - superb! Many thanks.
Glad it helped! You're welcome 💕💯🤩🙏😎
Xlog(-5)=log5: x=log5/log(-5): but -ve of log is undefined, as right side is +ve, left side should be +ve. If x is even no. And (-5)^x=5, it will true. So x=2×0.5=1 or 😅(5i)^x=5 : x=1/((logi/log5)+1)
but what is about the restrictions on the argument of logarithms. Can it be i?
Made it look very easy and the analysis is useful to remember. I like to call it "another tool in the math toolbox".
Cool, thanks🙏🙏🙏
This has nothing to do with Harvard. Harvard doesn't ask questions like this for deciding admissions.
No they don’t. They ask your gender and CRT questions.
@@saketashol6728no, not that either. They ask you, if you make the interview, what you see yourself doing there. What departments or professors drew your interest. How you will be of service to the community. Sometimes they WILL ask you about your particular work if you have anything in your background and file that is unusual. Like, you helped develop a program for treating cholera or your local church created a system to distribute clothing to the needy. Or you worked in a law office and studied some preliminary work. That’s what most of those interviews are like. Unless you’re a performing arts person and it’s graduate. That’s an audition.
(−5)^x = 5
x ln(−5) = ln(5)
x = ln(5) / ln(−5)
x = ln(5) / (ln(5) + ln(−1))
x = ln(5) / (ln(5) + πi)
What is the logarithm of a negative number ? Stop the crap
@@ioannisimansola7115 It is the logarithm of the positive number plus the logarithm of negative 1.
No solution is valid without testing on the equation. (-5)^x , x must be even . hence no solution.
Wrong.
@@Nikioko wright!
@@Nikioko
So, are you assuming that 1ⁿ=-1 does exist? Please, show me that n. I realy NEED to know that number makes 1ⁿ=-1 true.
@@ConradoPeter-hl5ij There is no real solution, but a complex one.
e^iπ = -1.
Work with that.
@@Nikioko
(-5)ⁿ=5
[(-1)×(5)]ⁿ=5
(-1)ⁿ×(5)ⁿ=5
(-1)ⁿ=5/5ⁿ
(-1)ⁿ=5¹-ⁿ
(-1)ⁿ×(-1)ⁿ=(5¹-ⁿ)×(-1)ⁿ
[(-1)ⁿ]²=(5¹-ⁿ)×(-1)ⁿ; but (-1)ⁿ=5¹-ⁿ
(-1)²ⁿ=(5¹-ⁿ)×(5¹-ⁿ)
[(-1)ⁿ]²=(5¹-ⁿ)²
[(-1)²]ⁿ=(5²)¹-ⁿ
[1]ⁿ=(25)¹-ⁿ
1ⁿ=25¹-ⁿ
but 1ⁿ=1 for every n.
Therefore,
25¹-ⁿ=1
25/25ⁿ=1
25=25ⁿ
5²=(5²)ⁿ
5²=5²ⁿ
2=2n
n=1 is a solution
but,
(-5)¹=-5
and
-5≠5
then n=1 is a absurd.
Therefore, do not exist a solution.
//////////////////////////////////////////////////
retake: (-1)ⁿ=5¹-ⁿ
if exist a n such that 1ⁿ≠1, so:
a=a (assuming this is true)
and a≠0
a/a=1
(a/a)ⁿ=1ⁿ
(a/a)ⁿ≠1
aⁿ/aⁿ≠1
aⁿ≠aⁿ
a≠a (a absurd conclusion)
when, a=0
0=0
0ⁿ=0ⁿ is true for every n≠0
when, n=0, n is not a incognite and 0ⁿ=0⁰ is indefinite.
Therefore, 1ⁿ≠1 is impossible.
(remember that is -1≠1)
//////////////////////////////////////////////////
Therefore, I can't see (-5)ⁿ=5 with a possible solution.
Thank you teacher!🎉
You're welcome 😊
INCOERENT.....
Thanks for the lesson
You're welcome! I'm glad you found it helpful. 🙏💕🥰✅
I will give you 5 marks out of 10. As you just only presented principle solutions. -1 = exp((2n+1)pi*i), where n belongs to Z (set of integers).
That's relevant here only if we consider -5^x to be a multi-valued function.
But i*2*pi*n plays a role in another part of the solution, which the author of the video missed:
exp(x * Log(-5)) = exp(Log(5))
x * Log(-5) = Log(5) + i*2*pi*n
x = (Log(5) + i*2*pi*n) / Log(-5).
And if we consider -5^x to be a multi-valued function, then the solution is:
exp(x * (Log(-5) + i*2*pi*m)) = exp(Log(5))
x * (Log(-5) + i*2*pi*m) = Log(5) + i*2*pi*n
x = (Log(5) + i*2*pi*n) / (Log(-5) + i*2*pi*m).
Log here represents the principal branch of the logarithm, i.e. Log(-5) = Log(5) + i * Pi.
For (2), that's not an answer, because we do not have an expression for log(i).
Muy bien explicado. Cuál sería el valor aproximado? es un número irracional?
Logarithms can interesting, learn to appreciate math education
So does thi mean that i*pi+ln(5) equals to lg(-1)+lg(5)? Based on the solutions you got.
Yes. The trick is change of base formula
@@superacademy247 Thanks, and also I forgot about the numerator.
Looking at the 2 solutions … 2* log = Pi ?🤔
Ιf e^i×π ις negative , how come it has a logarithm ?
MUUUITO BOMMM !!!
Solução bem interessante !
Thanks 😊💕🥰
SUPERB
I appreciate you watching! 👍🙏Thanks for the feedback! 🙏🤩
Why use log base 10 when you can use log base 5?
because log with base 10 is commonly used worldwide. when we write "log 2" this expression directly means that the base is 10. you can say we use it conventionally.
@@syther836 using log base 5 would significantly simplify the expression immediately, so using log 10 because it is used commonly worldwide is a lame excuse. Why not to add sin and cos there just because it is used worldwide?
You should have explain that first
you must know that i^2 is equal to -1 because i=√(-1) and √(-1)* √(-1) cancels out the √ and therefore gives you a -1.
The answer is incomplete because: e^(i(pi + 2*pi*n)) = -1
Pay attention, I shall be asking questions afterwards.
Incorrect solutions! There have not to be imagine numbers in denominator.
Hoc est x= 1/2, 1/4, 1/6, 1/8,......... Solutio infinitas. Responsi eheu!!!!
I know , i can equal -1 sub , but new at logarithms
Your pen is kind of dry and the way it scrapes across the paper makes the video a bit difficult to watch
The “Harvard University” headline is BS. He could say “MIT” or “Caltech” or anything else. The problem itself is uninteresting.
The easiest answer is X = 2/2 , because (-5) to the power of 2/2 is equal to √(-5) to the power of 2 is equal to 5
x = (2)^(1/2)
Solving one of those equations that would never occur in the real world.
I could, of course, learn how to do that and than keep and keep practicing it over my lifetime, just in case there's this 1 in a billion chance I will ever need it.
Or I could learn it once, when I actually need to in one of my next 10 lives for half an hour, to forget it then again thereafter because of how useless it is.
I'm all for math in applied settings, but endlessly solving pointlessly constructed, completely unrealistic equations, with methods that you'll have to practice all the time because you'll have forgotten half of it in 2 years, because that's how useful it actually is in life, just seems like playing sudoku as a hobby, rather than actually using math for something cool and real world.
Math practice that actually makes sense because you need it all the time, is to practice how to quickly calculate all kinds of things in your head, developing shortcuts to quick approximations and these kinds of things.
Solving unrealistic equations is just such a useless 'school skill' that it actually annoys me.
It's like a hobby, like playing sudoku, it doesn't make you smart and it doesn't give you any real useful skills. It just wastes your time. And if you stop practicing it for 2 years, you'll have forgotten half of it anyway, because you just never need it.
I love logarithms and logarithms love me
😂😂😂 Why didn't you end it at 9:12 but made another completely unnecessary step?
X=ln5/ln(-5)=ln5/(ln5+iπ)
lnE=log_eE
You seemed to have overlooked the solution of x=2/2, or even any even number over itself.
Thanks for sharing your perspective. 💯🙏💕🥰✅
Great job!
Thanks for the visit💪✅🥰
(-5)^x=5
X=1
:. (-5)^1 =5 X=1
You never really solved the problem, you just substitute for x = whatever
this whatever is the solution of the problem, whether you like it or not it will never change as will remain as it is now
3 months ago, I shared this video, but Nobody looked😢 it
х=-1.End! Vlad.A-Ata.
Impressive!! 🥀🥀🥀
Thank you! Cheers!🤩🤩🤩
@@superacademy247 How to verify the solution?
Gigo!
log i, is no solution...
Que questão bonita. Parabéns!
Thanks so much 💡😎🤩😍👏👏
Das muss doch viel einfacher möglich sein! 😅
You have lost x=2/2: (-5)**(2/2)=((-5)**2)**(1/2))=5 😂
crap! (-5)^a can be defined only for a = p/q (as f:R-->R)
(-5)^(sqrt(2)) = ??
ln(-1) does not exist! or == 0
2ln(-1) = ln(-1)^2 = ln 1 = 0 so ln-1 = 0 (if exists!)
so 0 = ln-1 = ln(e^(i*pi*x) = i*pi *x only for x=0 and this is not a solution!
@zenekk9684 He went over that. There can be no real solution, so we turn to Euler's identity. Is there some reason not to? Was it used wrongly?
Is there also no solution using log base 10?
I read your post three times and still don't know what you mean. To critique a demonstration, go step by step and make sure your notation is clear, just like in the video.
@@l.w.paradis2108
0 = ln 1 = ln(-1*-1) = 2 ln(-1) so ln( -1) = 0
-1 = e^(pi *i) in result
ln(-1) = ln(e^(pi *i)) = pi * i
something is wrong!
@@l.w.paradis2108
en.wikipedia.org/wiki/Complex_logarithm
@@zenekk9684 I thought that ln(a), where a < 0, does not exist in R. Zero is an element of R.
@zenekk9684 Okay, will look at that.
x = 2.5
😢konkret sadə cavab gözləyirdik.
Imaginario
Solution
Illiterate nonsense. One needs to specify the branch of the power function/logarithm before expression (-5)^x makes sense.
X=-1
👏 (-1)^(2n)=1
X=1
Glad I didn't go to Harvard.
You've lost a lifetime opportunity 🤗
-1
Bwamwabo mbuya mono genderera gokonya abanto baito
this looks super hard.
NOT really. Mastery of Euler identity you're good to go!
x is equal to an even number
Почему показываете ИНДУСОВ, как репетиторов? Они не умеют чётко говорить по английски!
But they can do an income tax return for you flawlessly 😂😂😂
boring
😴😴😴
So basically there is a solution group because we need to make it n* i* pi whereas the solutions makes no sense at all we put a rotation indicator in devisor . Nonetheless the analysis might be helping in a more complicated and realistic situation.
Thank you very very mach!!!🇮🇱🇮🇱🇮🇱
You're welcome 🤩🤩🤩
❤
FREE PALESTINE 🇵🇸🇵🇸🇵🇸
@@TommyBeaux от хамасовских бандитов!
@@leibmark
פלסטין חופשית מכלבה ישראלית כוס גדולה
-1
-1