x^(3x^2)=216

Solving a quadratic exponential equation for real numbers

[@markTheWoodlands]

This should be a PBS sponsored channel for children interested in math. As an added bonus atop the super clear delivery style - that hat really slaps.

[@edgardojaviercanu4740]

Children and not so children!

[@MacKo-gh6oe]

I love your videos, you are very articulate and explain math in a way that is simple to understand. Keep up the good work!

[@redroach401]

Using your video about the w lambert function, I was able to derive a general formula for x^^2 (a.k.a. x^x) = y and was able to solve this equation much faster, thank you soo much for your help

[@PrimeNewtons]

Excellent!

[@RexxSchneider]

It was going so well until around 8:20 when we get this misunderstanding: _"The actual answer is the positive version, not the negative version, because the negative version will give you imaginary numbers and remember what we want are just the real numbers_ (pointing to 'Find all real solutions' on the board)". The negative "solution" is a real number, so that's not the problem. The problem is simply that if you substitute -1.77 for x, you don't get 216, so the negative value _is not a solution at all._ I hope you can see the difference. Next we had a blatant error: "So if you have a negative number raised to a negative number for whatever reason as long as long as that negative number is not an integer you will always get imaginary answers." That's simply not true. Consider (-1/3)^(-1/3) = cube root of 3. I think we can agree that 1/3 is not an integer, nor is 3^(1/3) imaginary? Similarly for (-0.2)^(-0.2) or any number of the form -1/(2n+1) where n is a natural number. Finally, it's important when you're finding _all_ solutions to remember that the Lambert-W function is multivalued. Fortunately, all but the principal value of W(ln(36)) appear to be complex, -but I can't guarantee that there are no other real solutions.- Edit: I realised that the two values for branches 0 and -1 are sufficient when dealing with reals, and the latter turns out to be complex in this case.

[@ThenSaidHeUntoThem]

(-1/3)^(-1/3 is not real and that is not the cube root of 3 (-0.2)^(-0.2) is not real You wrote a lot to say at the end that all you claimed was wrong was actually right? What are you saying?

[@RexxSchneider]

@@ThenSaidHeUntoThem Do you understand what real means? (-1/3)^(-1/3) = (-3)^(1/3) = cube root of -3 which is approximately -1.442. What makes you thinks that's not a real number? (-0.2)^(-0.2) = (-1/5)^(-1/5) = (-5)^(1/5) = fifth root of -5 which is approximately -1.380. Also real. What I claimed is right. You're talking out of your backside. Want to try again?

[@wes9627]

Fixed-point iteration is the simplest solution method: 3x^2*lnx=ln216; x=√[(ln216/3)/lnx] x←√[(ln216/3)/lnx] yields x=1.7707... A negative x would require 3x^2 to be a multiple of 2, which would not satisfy the given equation.

[@blackdye2420]

New Brazilian follower here!

[@rakeshsrivastava1122]

Very well explained. You connect with the viewers very well.

[@Katavicta]

My dawg just snuck in the sqrt... lol

[@Sigma.Infinity]

Interesting video. On a pedantic note, 8:26, a negative number raised to a negative rational number with an odd denominator will have a real solution. (Doesn't apply to the present problem.)

[@rca7591a]

I've always known math can be almost unimaginably complex, but I've seen some shit on this channel I've never dreamed of imagining. It just keeps getting more and more interesting. 😎😎

[@kingbeauregard]

This channel delights and disturbs me. Only the hats keep me calm.

[@rca7591a]

Rewriting the problem. My calculus teacher in college was big on rewriting a difficult problem... Completing the square, Making some thing equal to 1 and such like strategies in problem solving. Mr Harold Line was his name. He's likely gone now. Claimed by time and history now, as it does us all. I remember solving many a long equation by changing fractional terms to inverse powers and multiplying. The most significant methodology of problem solving I remember is the inverse power rule of antiderivatives. That made a lot of things calculus very clear to me. When I think about a problem or look at an equation I still hear your voice and what you taught Mr Harold Line. God be with you sir. I see and hear a lot of him in you Primenewtons. Mathematics is indeed a language all in it's own. 😎😎🙏🙏

[@oluwaseunakinte5184]

Nice one, please represent the square root very well, excellent teaching from you.

[@Imposter7777]

Lol dont think I didnt notice that fade in at 7:52

[@sunil.shegaonkar1]

Anyone please find the error: I jump at 3:05: where x^x = 6^(1/2); after following power & reduction, => x= 6^(1/6), which does not match with answer.

[@baselinesweb]

No problem ... but there's just one problem LOL - love it. Almost as much as I love that magic /2 that appears. Good job. I almost got through it myself, but now I need to go refresh that W stuff in my head. Thank you.

[@RandomGuyEdit]

Please tell a method to calculate the W of integers manually ... It will be really helpful ... Please ..

[@RexxSchneider]

You can't express the Lambert W in terms of elementary functions. There are series expansions for W(x) and there are integral representations. None of these are particularly easy to evaluate manually. Wikipedia has a lengthy article on the Lambert W function. You can get numeric approximations from Wolfram Alpha which calls it "ProductLog(k, x)" where k is the branch (0 for principal branch) and x is the input. For example ProductLog(0, ln(36) gives the result 1.1428340634250633101669478241795999773625132790388322670904242154 which is good enough for most applications!

[@necrolord1920]

I didn't see the solution at first, but when you got to 5:07 I had the aha moment. Love the video.

[@bartholomewrichards1663]

Enjoyed yer style and info.very useful

[@fsisrael9224]

Flower? I only know about the fish that if you multiply it by e raised to the power of the same fish and use the w function, you get the fish back

[@justekiara1953]

What a performance in maths. Thank-you.

[@ranitchatterjee5552]

How did you choose the value of W?

[@BlueSiege01]

amazing video!!!!

[@dinabandhusaha5520]

Please make a video on Newton's method

[@PauloDacosta-s1s]

Why did you developed the equation and did not apply the productlog in the original equation? It was very clear since the beginning that you may need to use the productlog so it will be easier to reshape the original equation to use productlog….or I am wrong?

[@yigitgumus09]

how a glorious video

[@SeC0nD_ChAnCE]

hey in this math prb cant we use a to the power a equalsb to the power b then a=b? that way we get x= square root 6??

[@anestismoutafidis4575]

2^(3•2^2)=4096; log (216)=2,334; (216)^1/6=2,4494 2,4494 - 2,3344=0,115 (0,115)^1/6=0,697 2,449•0,697=1,707 1,707^(3•1,707^2)=107,17 1,77^(3•1,77^2)=214,10 1,77078^(3•1,77078^2)=216,0 x=1,77078

[@surendrakverma555]

Very good. Thanks Sir

[@renardtahar4432]

very nice master!

[@วิระพิทักษ์ถิร]

This problem solving consists of an analysis solution and a numerical solution .

[@levskomorovsky1762]

From x^x^2 = 6: x^2 lnx = ln6; ln x e^ ln (x^2) = ln6; ln x e^2 ln x =ln 6; 2ln x e^(2 ln x) = 2 ln 6; W (2ln x e^(2 ln x)) = 2ln 6; 2 ln x = W (2ln 6); ln x =[ W (2 ln 6)]/2 Minus is dissapeared! Is it right?

[@DefenderTerrarian]

That gives you the real solution, to get the imaginary solution, you need to consider ln in the imaginary world as well. However, that is a very neat solution as well.

[@minhhieunguyen7638]

Can you help me x^(1-x) - x = 32? Thank you so much!

[@samirayoub6100]

No solution, if you use the comparative graphs of f(x)=x^(1-x) - x and g(x) =32, they will never intersect. For any value of x, f(x) is less than 32

[@glennjohnson4919]

Aren’t there three cube roots of 216?

[@glennjohnson4919]

Just real, missed it.

[@panjak323]

I got stuck at t^t = 6^2, Don't know what W function was supposed to mean

[@Bertin-q3y]

Une solution unique > e^(-0,5)

[@tamilselvanrascal5956]

🎉🎉🎉

[@ganda3454]

x^3x^2=6^3 x^x^2=6

[@samirayoub6100]

very nice, but if you introduce the ln since x^x^2=6, you have only the positive result

[@pianisissimo4459]

He explain It in the video

[@susanalabbe2433]

wow bravissimo

[@emmanuelseiman2725]

You forget the root in the last line.

[@goodboyneon]

He divided the exponent by 2, which is the same as taking the square root

[@BlueSiege01]

Yes

[@TomLeg]

What is W(ln 36) ? i.e ln(36) / e^ln(y) age 68 so not a child :-)

[@tatuvedovello]

English speakers need to understand that using the W function is NOT solving the problem

[@binaryblade2]

That's like saying an answer in terms of cosine is not an answer.

[@JohnGavain]

How could I have known that I had to sqare both sides in the step that yields the 36 as a result? Guessing it? I think math is sometimes indeed to hard for me. Or haven't I seen enough of these examples to be capable of guessing the right calculation path? Smh!

[@theupson]

this kind of fiddling is necessary in the olympiad versions of this problem, where the equations can always be finessed into A^A = B^B. i imagine that kind of primed OP, but for this problem if your first thought is "W", it works fine to take log of both sides as your first step. the work proceeds very like to the way the video progresses after the log.

[@kingbeauregard]

BOOOO! Lambert W function never feels like a solution to me. I had a thought though. if t^3 = 216, it doesn't mean that t must be 6. Rather, (t - 6)*(t^2 + 6t + 36) = 0, which allows for two other solutions. As it turns out, those other solutions aren't real, so we can discard them. But, it's good to be aware of them.

[@RexxSchneider]

If t^3 = k, then t = k^(1/3) or k^(1/3)*(-1 + i√3)/2 or k^(1/3)*(-1 - i√3)/2. The values (-1 ± √3)/2 are the two complex cube roots of unity, so it's trivial to find the three complex cube roots of any constant.

[@redpepper74]

The thing about some of these exponential equations is that they often don’t have algebraic solutions. That’s why we have to use the product log function, because it gives us access to those values we can’t express otherwise.