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MathProwess.
Приєднався 3 кві 2024
Hi, learn math of all levels ranging from elementary to advanced in a simplified way ✍️.
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How To Solve x=In(x) | Can You Solve This Logarithmic Equation?
Hello my great people, Today's math's tutorial is on logarithm.
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Watch and learn something new please.
Subscribe my channel.
Like, comment and share please 🙏🙏🙏
Thanks a million.
#mathprowess #logarithm #education #maths #solve
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How To Solve Sqrt(Inx)=In∛x | Math Olympiad Problem
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Algeria | Algeria Math Olympiad Problem
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Olympiad Math Problem
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Olympiad Math Problem, is a must watch math video tutorial for every math students who want to excel in the world of mathematics. Watch this video from the beginning to the end with skipping any parts for a better understanding. Leave a comment in the comments section. Like and share this video clip via the link here; ua-cam.com/video/K_kjprmP2So/v-deo.html Thanks for stopping by 🙏🙏🙏 #algebra #...
Math Olympiad Problem | Solving A Nice Radical Equation.
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VietNam || VietNam Can You Solve This Math Olympiad Problem?
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What Is The Square Root Of Iota(i)? | How To Find The √ i.
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A Beautiful Algebra Math Solution
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India | Which Is Greater Between 1000^1001 Vs 1001^1000 ?
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Limit Of A Function Using Taylor's Series | Taylor's Series
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University Entrance Math
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Impossibility Made Possible With Math World
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This problem appears to be so difficult but thanks to mathematics world where there is no problem without a solution. A world where impossibilities are made possible with reasonable proves Enjoy watching this video clip and learn some unique skills from our channel. Like, and comment. Share via the link here below; ua-cam.com/video/ztF0r8a9GIs/v-deo.html Subscribe our channel for more math skil...
What Is The Remainder?
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Viral Math Olympiad Problem
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Trending Exponential Equation | Why This Math Video Is Trending.
Oxford Entrance Algebra Problem For Master Degree Students.
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Olympiad Mathematics | Solving A Radical Equation With Imaginary Number
I am in 7th grade about to do algebra 2 any tips?
2 real solutions, because -1/e < -ln4/5 < 0 1.5271834618689137843439681976847183206615191189475680905718280703... or 7.0378231590674309036420394728342539963399273768491109730261293405...
No real solutions
Only complex: 0.063959810301343172499210859310985355287605730479104740675403973... - 1.0908433765399575763735988438001669099914583549817516466302112... i or 0.063959810301343172499210859310985355287605730479104740675403973... + 1.0908433765399575763735988438001669099914583549817516466302112... i or 1.248405605048936531526278135104793156575121368698758722353396678... - 5.504574134828021237818776295401377186833156083716398821615796055... i or 1.248405605048936531526278135104793156575121368698758722353396678... + 5.504574134828021237818776295401377186833156083716398821615796055... i or ...
Kudos man...😂😂😢
{4x+4x ➖ }{9x+9x ➖ }={8x^2+18x^2}=26x^4 2^13x^4 2^13^1x^4 2^1^1x^2^2 1x1^2 x^1^2 (x ➖ 2x+1). {6x+6x ➖ }=12x^2 3^4x^2 3^2^2x^2 3^1^1x^2 3x^2 (x ➖ 3x+2).
Muy sencillo, X =5, Y= -6.
This beautiful man. Keep running this channel sir and the sky is yours.....❤
👍👍
Cool one 😅😅
Cheap proof.
I pove this in my head....😂😂😂
Pove in your head??
Nice work
Thanks for the visit
Nice proof man.
Thanks!
+1 or -1
That is just two answers out of the 16 roots to this math problem sir, kindly watch the full video for the rest 14th roots which can not be found on the surface. Thanks sir.
Where do you get x_1 = sqrt(2+sqrt(2))/2 + (sqrt(2-sqrt(2))/2 * i from? Same with the other x_(odd number). Not explained. Hint: use the half-angle formula.
every x^n equation will has n roots 😅😅😅 so x^16 -->16 roots 😅😅😅
Yes,
+_1, if x is from R.
Correct sir.
@@Mathprowess Thank You!
Geez, all you really have to do is draw the circle r=1 on the complex plane. You know there are 16 roots. So the circle is 2pi/16. Then just cos +isin of all those angles.
Fastest and easiest way, thanks for suggesting this approach
I have never been so lost in my life...
How do you mean?
The moment you introduces Euler's equation I wasn't able to follow anything you said, surely there's an easier method to solve this...
Doubt this is on an Oxford entrance paper. Basic application of cos(theta)+isin(theta) and knowledge of de Moivre’s theorem.
Ok
Well detailed, thanks.
Wow! I have learnt a lot from your channel since i subscribe. Thanks sir ❤❤❤
I'm glad you did
16 log x = log 1 = 0 log x = 0 x= 1
You shouldn't assume x > 0, though
not to mention you can also change side by making it 16 root 1 which is still = to 1
Short cuts kill faster than the metal bullet in the world of math.
M=n, simple.
So what then is the numerical value to m ?
😂😂❤❤
X^16 = 1, 1^2 = 1, 1^4 = 1, see where this is going ?
Alright sir.
This is a 30 seconds challenge: 3^3 == -1 mod 7, therefore 3^333 == -1^111 mod 7, 111 == 1 mod 2, therefore 3^333 == -1 mod 7. 7-1 = 6. The remainder is 6.
Master
Why does -1 ^ 2n equal 1? Cant we just do -1^2
Nope
Why this is still on yt? Please remove it unless you will gain a lot haters
I will shortly sir. Thanks for reminding me sir.
Great tutorial man
You nailed it on the head here. In fact, this is one of the best math videos from this channel so far.
Wow, thanks!
This is wonderful, great step man. Thanks for tutorial sir....❤❤❤
Spoilers (potentially I haven’t watched the vid yet) Wlog assume m>=n 1/m <= 1/n 1/5 = 1/m+1/n <= 1/n + 1/n = 2/n 1/5 <= 2/n, 5>=n/2, 10>=n 1/5=1/m+1/n>1/n 1/5>1/n 5<n 5<n<=10 Now just check all 6 possibilities. n=6 gives m=30 n=7 gives no solution n=8 gives no solution n=9 gives no solution n=10 gives m=10 So solutions (m,n) are (5,30) (10,10) and (30,5)
Stupid leanthy answer।। X^1/x= (- 1/3)-3
I still love you for the insult sir.....😍😍🤣🤣😂
Simple problem.
Nice work sir.
3^3 (x ➖ 3x+3) .
Good day sir, I seem not to get your point here sir. Is this a question or the procedure to this challenge?
-1/3, I solved it in 3seconds in my head 😂😂😂
You the master here then...
its obvious that x=1 and -1. For what reason this task?????
How do you get the other 14 roots? That is just the reason for this task sir. Thanks watching and commenting sir.
1 and -1 are the two real roots out of the total 16 roots. The rest of them are imaginary. The task was to find all the roots, not only the ones who become to the R
X⁴+4=0> ,(X²)²+ 2²=0,(x²+2)(x²+2)=0,x²+2=0,x²=-2, la somme de deux Carré n'est jamais nul,S={0}.
This approach will give only two roots out of the four roots we are expected to get from here sir. Thanks for the suggestion sir.
Who cares? This took me about 30 seconds using Windows Calculator.
#icare #stopbullying #nomorehate
But the question suggest you must not use the calculator.
Again if you do not care others care...😍😍😂🤣🤣
My solution: x^x = x^2 X ln x = 2 ln x X ln X - 2 ln X = 0 (X - 2)(ln X) = 0 X - 2 = 0 X = 2 Ln X = 0 X = 1
You are great sir. Thanks for sharing your approach with Onlinemaths TV
The first 8 solutions only required me to abuse the difference of two squares. However, once I began contemplating the breakdown of x⁸ + 1, I realized that a conversion to polar coordinates might be helpful.
You cannot have 20 roots for a 16 degree polynomial unless you are operating outside of the complex numbers. Either some of your roots are the same or you made an error.
I would sketch it on the complex plane. Makes it clearer where the 16 solutions are.
That is another wonderful approach also sir.
Решается легко. Надо на комплексной плоскости начертить единичную окружность и разделить ее на равные 16 частей. Каждая точка на такой трудности есть решение данного уравнения (комплексное число, включая частные решение такое как: ±1 и ±i)
Noted sir, I will draw the unit circle for clearer picture of the solutions in subsequent problems. Thanks for the suggestion sir.
I have solved something similar. 😊❤
sure buddy
Great job well done...❤
Thank you! Cheers!
this made my opinion on oxford drop drastically
I agree. The video should've ended just before throwing in sines and cosines. The exponential conveys the 16 roots of unity so succinctly and I can't imagine why an interviewer would want to see any more than that. On top of that I'm not really sure what kind of interview question is, it's not really approachable. More like a you know it or you don't kind of problem.
@@drkn0ckers715 i meant, this is what we've been doing in 1st year of engineering at a college that just isn't all that prestigious. i can't possibly imagine it's supposed to be a problem for anyone trying to get info oxford for a masters degree
this is not a actually an interview question for oxford maths. it may possibly be the question asked in the first 2 minutes of an hour long interview for oxford maths.
@@drewkavi6327 the fact that it's the first of many doesn't make it any less of an interview question lmao
@@blue1580 i mean, it is the kind of question that is just there so the applicant feels an ease before actually being challenged. every applicant will be expected to know this obviously
👍