Solving x^{ln(x)}=xln(x) | A Log Equation

[@Mediterranean81]

5:00 i would subtract t on both sides and differentiate to get 2t-2=ln(t) Then exponentiate both sides So (e^2t-2)=t Differentiate with respect to t and simplify to get 2(t-1)e^2(t-1)=e Take the Lambert W function 2(t-1)=W(e)=1 t-1=1/2 t=3/2 x=e^3/2 and x=0 are the solutions

[@stvp68]

In music, tutti (fr Italian) means “everyone” - so we’re all in included!!!

[@DrR0BERT]

How about showing f(t) = g(t) only happens when t = 1? Consider h(t) = f(t) - g(t). h(t) is continuous on (0,+∞), with derivative h'(t) = 2t - 1 - 1/t. If t = a is a second solution, then by Rolle's Theorem there would be a point t = b in (0,+∞) strictly between 1 and a with f'(b) = 0. There is only one solution to h'(t) = 0 in (0,+∞): t = 1. So no b exists. Therefore the only solution to f(t) = g(t) is t = 1, which is what you wanted.

[@miguelcerna7406]

Amazing. Where do you find these problems? There is just something ridiculously beautiful about the "tangency" between both functions being precisely at the point (e,e) F-ing so cool! lol

[@franckplanks]

Solved by inspection.

[@christopher679]

Yup

[@bramont6225]

Gracias profesor bonito ejemplo

[@conradosetti3781]

Does anyone know what sofware he uses to make his black boards

[@relentlessmind9008]

Thank you so much! На русском спасибо - Just wanted to tell you thanks for your effort. Also wanted to ask you proff, can you add some trig equations from time to time? It would be awesome!

[@mehrdadbasiri9968]

Nice one... Thanks for your videos...👍👍👍.

[@Blaqjaqshellaq]

Unlike f(t) and g(t), f(x) and g(x) "osculate" (kiss?), because f''(x)/g''(x) is positive while f''(t)/g''(t) is negative.

[@alphaomega3944]

Math class AND art class. Cool indeed.

[@Paul-222]

This was trickier than it looked at first.

[@stevemonkey6666]

If it is confusing to not have parentheses then should you not put them there in the first place? Seems weird to me that you started your video by putting the parentheses in, when you should have put them in originally. And by this I am referring to lnx^x

[@nasrullahhusnan2289]

RHS of the equation should be interpreted as ln(x^x,) or [ln(x)]^x? To distinguish the two how we must write them without using bracket?

[@SyberMath]

It’s ln(x^x) because lnx is not in parantheses

[@nasrullahhusnan2289]

I don't get your explanation due to the my unability to write properly exponential with variable. Anyway thanks for the response. To make it clearer I'll use number as exponent. How we write ln(x²) and [ln(x)]² with no bracket so that it is clear which expression we mean. From your explanation we write the first as lnx² since no bracket is needed in writing ln. The second one must be written as (lnx)². No way not to use any bracket.

[@Packerfan130]

x^(ln x) = ln x^x x^(ln x) = x ln x Note that if 0 < x < e, LHS > 0 but RHS < 0 so no solution on (0, e) ln x^(ln x) = ln (x ln x) (ln x)*(ln x) = ln x + ln (ln x) (ln x)^2 - ln x = ln (ln x) (ln x)(ln x - 1) = ln (ln x) Note that ln (ln e) = ln 1 = 0 and (ln e)(ln e - 1) = (1)(1 - 1) = 1*0 = 0. Thus, x = e is a solution. Define f(x) = (ln x)(ln x - 1) - ln (ln x) for all x > e Critical points where f'(x) = 0 f ' (x) = (1/x)(ln x - 1) + (ln x)(1/x) - 1/(ln x) * 1/x f ' (x) = (2/x) ln x - 1/x - 1/(x ln x) f ' (x) = 0 when (2/x) ln x = 1/x + 1/(x ln x) 2 ln x = 1 + 1/(ln x) 2 (ln x)^2 = ln x + 1 2 (ln x)^2 - ln x - 1 = 0 ln x = (1/4)(1 +/- sqrt((-1)^2 - 4*2*-1))) ln x = (1/4)(1 +/- sqrt(9)) ln x = (1/4)(1 +/- 3) ln x = 1 or ln x = -1/2 x = e or x = e^(-1/2) However, both are outside the domain of f(x). Thus, no solution for (e, infinity)

[@broytingaravsol]

x=e

[@elenadraganova7992]

My way: x^(lnx)=xlnx x^(ln(x/e))=lnx , ln both sides ln(x/e)lnx=lnx lnx=0 or ln(x/e)=1 X=1 or X=e^2

[@jeremygalvagni9266]

You wrote ln both sides, but didn't then ln the right. ln(x/e)lnx=ln(lnx)

[@giuseppemalaguti435]

Io ho fatto così... Applico ln... lnxlnx=lnx+lnlnx... pongo lnx=t.. t^2-t=lnt..una soluzione è t=1,cioe x=e, che dovrebbe essere unica ah ah

[@mychmose]

I love your videos.

[@scottleung9587]

Nice!

[@matematikguzeldir]

I've watched this video last month @mathpositive channel. I guess you steal other's questions.

[@chrisjuravich3398]

Because these questions are in the public domain, it’s not stealing. I think it’s very helpful to see how other people work the same problem.

[@matematikguzeldir]

@@chrisjuravich3398 At least he could have said the source. Of course we all solve similar questions but this one is an equation written by @mathpositive. So a quotation would be fair

[@nodawg]

nice vid

[@james10492]

Interesting problem

[@barakathaider6333]

👍

[@leonardobarrera2816]

just a perdect video

[@souzasilva5471]

De onde saiu o igual a zero? Where did the equal to zero come from? Resposta forçada.

[@peteneville698]

Why not just say "log" instead of saying the letters "l" and "n" - it's so annoying. Everyone knows that "ln" is the natural log - absolutely no-one is going to assume base 10 just cos you said "log".

[@kianmath71]

X = e

[@rakenzarnsworld2]

x = e