Harvard University Admission Interview Tricks

[@andretewem3385]

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.

[@СенчуринНиколай]

bravo!

[@216dark2]

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.

[@andretewem3385]

What symbol do you not understand? Often I write Ln for the complex logarithm and ln for real logarithm (normal).

[@216dark2]

@@andretewem3385 "^^", what does it mean? 😅

[@216dark2]

@@andretewem3385 "^^" power of 2?

[@LuisGrandeGomez]

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.

[@adamkolany1668]

@0:5:33 you may not use log in complex domain in the way you did it here.

[@sanjitkumarsinghmaimom4846]

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)

[@rayrocher6887]

I recommend,-1, sub set, thanks for the math lesson, try hard at thinking, encouragement, great challenge test

[@DarthQuantum-ez8qz]

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).

[@letsimage]

but what is about the restrictions on the argument of logarithms. Can it be i?

[@vencik_krpo]

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.

[@Penndennis]

That was great Man! Two methods; very well explained - superb! Many thanks.

[@superacademy247]

Glad it helped! You're welcome 💕💯🤩🙏😎

[@tysonsmat9918]

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).

[@denvned]

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.

[@KipIngram]

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]

You forgot some stuff: ln(−5) = ln(5) + ln(−1)= ln(5) + πi Therefore, x = ln(5) / ln(−5) = ln(5) / (ln(5) + πi).

[@KipIngram]

@@Nikioko Ah, good point. Thanks.

[@robaxhossain5653]

@@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.

[@CuriousCyclist]

Thank you for taking the time to make this video. Much appreciated. ❤

[@superacademy247]

You are so welcome! Glad it was helpful. 🙏🙏🤩🤩

[@mouradbelkas598]

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

[@stardustwight1895]

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.

[@mouradbelkas598]

@@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

[@carloscifuentes5091]

Muy bien explicado. Cuál sería el valor aproximado? es un número irracional?

[@X00000370]

Made it look very easy and the analysis is useful to remember. I like to call it "another tool in the math toolbox".

[@superacademy247]

Cool, thanks🙏🙏🙏

[@dumitrudraghia5289]

INCOERENT.....

[@rayrocher6887]

Thanks for the lesson

[@superacademy247]

You're welcome! I'm glad you found it helpful. 🙏💕🥰✅

[@zsombororovec645]

So does thi mean that i*pi+ln(5) equals to lg(-1)+lg(5)? Based on the solutions you got.

[@superacademy247]

Yes. The trick is change of base formula

[@zsombororovec645]

@@superacademy247 Thanks, and also I forgot about the numerator.

[@lourdesvillamayor-nu5ld]

Thank you teacher!🎉

[@superacademy247]

You're welcome 😊

[@hokie6384]

Looking at the 2 solutions … 2* log = Pi ?🤔

[@ssalmero]

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.

[@Nikioko]

(−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)

[@ioannisimansola7115]

What is the logarithm of a negative number ? Stop the crap

[@Nikioko]

@@ioannisimansola7115 It is the logarithm of the positive number plus the logarithm of negative 1.

[@rayrocher6887]

Logarithms can interesting, learn to appreciate math education

[@ioannisimansola7115]

Ιf e^i×π ις negative , how come it has a logarithm ?

[@syedmdabid7191]

Hoc est x= 1/2, 1/4, 1/6, 1/8,......... Solutio infinitas. Responsi eheu!!!!

[@gibbogle]

For (2), that's not an answer, because we do not have an expression for log(i).

[@pelasgeuspelasgeus4634]

😂😂😂 Why didn't you end it at 9:12 but made another completely unnecessary step?

[@jtinalexandria]

This has nothing to do with Harvard. Harvard doesn't ask questions like this for deciding admissions.

[@saketashol6728]

No they don’t. They ask your gender and CRT questions.

[@wideeyedraven15]

@@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.

[@vfa1985]

SUPERB

[@superacademy247]

I appreciate you watching! 👍🙏Thanks for the feedback! 🙏🤩

[@mouradbelkas598]

No solution is valid without testing on the equation. (-5)^x , x must be even . hence no solution.

[@Nikioko]

Wrong.

[@atheroot]

​@@Nikioko wright!

[@ConradoPeter-hl5ij]

​@@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.

[@Nikioko]

@@ConradoPeter-hl5ij There is no real solution, but a complex one. e^iπ = -1. Work with that.

[@ConradoPeter-hl5ij]

​@@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.

[@claudiohase296]

MUUUITO BOMMM !!! Solução bem interessante !

[@superacademy247]

Thanks 😊💕🥰

[@joeaberman449]

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.

[@ВикМитов]

x = (2)^(1/2)

[@bumbarabun]

Why use log base 10 when you can use log base 5?

[@syther836]

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.

[@bumbarabun]

@@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?

[@jeffreyluciana8711]

I love logarithms and logarithms love me

[@atheroot]

Incorrect solutions! There have not to be imagine numbers in denominator.

[@giuseppemalaguti435]

X=ln5/ln(-5)=ln5/(ln5+iπ)

[@ВасильМигович-ш5п]

The answer is incomplete because: e^(i(pi + 2*pi*n)) = -1

[@xzxz214]

The “Harvard University” headline is BS. He could say “MIT” or “Caltech” or anything else. The problem itself is uninteresting.

[@SGuerra]

Que questão bonita. Parabéns!

[@superacademy247]

Thanks so much 💡😎🤩😍👏👏

[@Maths__phyics]

3 months ago, I shared this video, but Nobody looked😢 it

[@hustledude]

Your pen is kind of dry and the way it scrapes across the paper makes the video a bit difficult to watch

[@rayrocher6887]

I know , i can equal -1 sub , but new at logarithms

[@Gwynbuck]

Pay attention, I shall be asking questions afterwards.

[@surinetso8346]

(-5)^x=5 X=1 :. (-5)^1 =5 X=1

[@BruceLee-io9by]

Great job!

[@superacademy247]

Thanks for the visit💪✅🥰

[@skyrubber]

lnE=log_eE

[@kel1404]

You never really solved the problem, you just substitute for x = whatever

[@syther836]

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

[@Пщдпку]

х=-1.End! Vlad.A-Ata.

[@regularguy9264]

You seemed to have overlooked the solution of x=2/2, or even any even number over itself.

[@superacademy247]

Thanks for sharing your perspective. 💯🙏💕🥰✅

[@HoWong-g4o]

x = 2.5

[@andreykloubovich892]

You have lost x=2/2: (-5)**(2/2)=((-5)**2)**(1/2))=5 😂

[@brainard30]

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

[@kvadromir]

👏 (-1)^(2n)=1

[@sonnyandersonmanu5958]

log i, is no solution...

[@peterotto712]

Gigo!

[@l.w.paradis2108]

Impressive!! 🥀🥀🥀

[@superacademy247]

Thank you! Cheers!🤩🤩🤩

[@BacLe-r9f]

@@superacademy247 How to verify the solution?

[@จารุญ-ย7ข]

X=-1

[@gulleylazeynalova4928]

😢konkret sadə cavab gözləyirdik.

[@Luis-lm2lg]

Imaginario

[@superacademy247]

Solution

[@ff7113]

-1

[@gulleylazeynalova4928]

X=1

[@zenekk9684]

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!

[@l.w.paradis2108]

@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.

[@zenekk9684]

@@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!

[@zenekk9684]

@@l.w.paradis2108 en.wikipedia.org/wiki/Complex_logarithm

[@l.w.paradis2108]

@@zenekk9684 I thought that ln(a), where a < 0, does not exist in R. Zero is an element of R.

[@l.w.paradis2108]

@zenekk9684 Okay, will look at that.

[@MartinJefferies-j1d]

Glad I didn't go to Harvard.

[@superacademy247]

You've lost a lifetime opportunity 🤗

[@xgx899]

Illiterate nonsense. One needs to specify the branch of the power function/logarithm before expression (-5)^x makes sense.

[@justinokenye8116]

Bwamwabo mbuya mono genderera gokonya abanto baito

[@astropatroldc]

x is equal to an even number

[@richardmullins44]

this looks super hard.

[@superacademy247]

NOT really. Mastery of Euler identity you're good to go!

[@samwi-fifi1120]

😴😴😴

[@gibson2623]

boring

[@nikolairomanov7509]

Почему показываете ИНДУСОВ, как репетиторов? Они не умеют чётко говорить по английски!

[@KasumiModa]

But they can do an income tax return for you flawlessly 😂😂😂

[@winniethexiinwesttaiwan8578]

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.

[@אריהקופרמן-ל7מ]

Thank you very very mach!!!🇮🇱🇮🇱🇮🇱

[@superacademy247]

You're welcome 🤩🤩🤩

[@leibmark]

❤

[@TommyBeaux]

FREE PALESTINE 🇵🇸🇵🇸🇵🇸

[@leibmark]

@@TommyBeaux от хамасовских бандитов!

[@adventure9544]

​@@leibmark פלסטין חופשית מכלבה ישראלית כוס גדולה

[@delciocandido1024]

-1

[@agrolactingenieriadealimentos]

-1