Solving the Exponential Equation xe^{1/x}=e

[@lambda2693]

Lambert log successfully gives the correct answer, so I think it’s worth a mention

[@chaosredefined3834]

Let's make it more general x e^(a/x) = b, where a, b is some constant. Take the reciprocal of each side 1 / (x e ^(a/x) ) = 1/b Tidy things up a bit (1/x) e^(-a/x) = 1/b Multiply each side by -a (-a/x) e^(-a/x) = -a/b Take the Lambert W function of each side W( -a/x e^(-a/x) ) = W(-a/b) By definition, the LHS = -a/x -a/x = W(-a/b) Take the reciprocal of both sides -x / a = 1 / W(-a/b) Multiply both sides by -a x = -a / W(-a/b) Enjoy.

[@walterufsc]

Solution without using logarithms: x * e^(1/x) = e Raising both sides to the x power: x^x * e = e^x Hence: e^x / x^x = e => (e/x)^x = e = (e/1)^1 => x = 1

[@threstytorres4306]

Using the Lambert W Function where W(xe^x) = x xe^x = 1/e => W(xe^x) = W(1/e) => x= W(1/e)≈ 0.28

[@abhinavbhutada9b484]

Thumbnail and question different actually

[@SyberMath]

No more 🫣

[@NikolayVityazev]

Good!

[@gdtargetvn2418]

The thumbnail is incorrect.

[@SyberMath]

I just realized that and fixed it! Sorry

[@giuseppemalaguti435]

x=-1/W(-1/e)=1

[@andreabaldacci1142]

I set e^(1/x)=u, so that x=1/lnu. The initial equation becomes u/lnu=e, so u=elnu e u=ln(u^e). This is equivalent to e^u=e^ln(u^e), or e^u=u^e. Hence u=e, or e^(1/x)=e, which gives x=1.

[@TorAulin]

Very simple!

[@noahvale2627]

I like your tables. Actually, I like seeing tables, equations and graphs so all three aspects can be seen.

[@SyberMath]

Glad to hear that! 🤩

[@veganforanimals418]

I differentiated both sides and got x=1 , can exponential equations be solved by differentiating

[@tbg-brawlstars]

X = W(1/e)..... Bruh

[@leonardobarrera2816]

I got an irrational answer with thi w lambert function

[@ssalmero]

inverting the equation and negating we get: -(1/x).e^(-1/x)=-e^-1. Taking Lambert's x=1

[@d_mcg]

x e^{1/x} = e let y = 1/x x = 1/y (1/y) e^y = e e^y = e y Divide both sides by e: y = e^(y - 1) WolframAlpha is then able to reduce this to: y = -W(-1/e) y = 1 But I wasn't sure how WolframAlpha got it into Lambert W function form. Working backwards: Negate both sides: -y = W(-1/e) Undo the Lambert W function: W(-y e^ -y) = W(-1/e) -y e^ -y = -1/e Negate both sides again: y (1/e^y) = 1/e Take the reciprocal of both sides: (1/y) e^y = e Which is the form we had it in earlier on line 4.

[@kurtdobson]

Using Matlab Symbolic Toolbox (so my brain doesn't hurt): tic; clear all tic syms x eqn = (x*2.71828)^(1/x) == 2.71828 [x] = solve(eqn,x); (simplify(x,'steps',50)) toc ans = 1 Elapsed time is 1.140588 seconds.

[@VuAilleurs]

x=1 finished.

[@Rbmukthegreat]

Using lambert W, 1/(xe^{1/x})=1/e -1/x * e^{-1/x}=-1/e -1/x = W(-1/e) x = -1/W(-1/e). So we also derived that this number is 1!

[@renesperb]

It took me 2 seconds to find solution .

[@mariomestre7490]

Genial: l'ajuda de les funcions

[@moeberry8226]

Another problem solved by the Master.

[@SyberMath]

Thank you! 🥰

[@moeberry8226]

@@SyberMath no problem

[@ErichEliskases-t3j]

thought it's just writen as : y'+y=1 . and the sol is obvious.

[@mcwulf25]

Easy to find the answer but the real maths is in showing it's the only answer. (It usually is with these exponentials)

[@lgooch]

I just got (-1)/(w(-1/e))

[@Jack_Callcott_AU]

@l’gooch That is what i got too. I better see what Wolfram Alpha says.

[@itzrealzun]

Preview has a mistake

[@SyberMath]

I just realized that and fixed it! Sorry

[@armacham]

take the natural log of both sides, then move things over to get: 1/x = 1 - lnx make a new variable u u = 1/x u = 1 - ln(u^-1) u = 1 + lnu u - lnu - 1 = 0 define the function f(u) = u - lnu - 1 f'(u) = 1 - 1/u f'(u) will be equal to zero only when 0 = 1 - 1/u 1 = 1/u u = 1 this corresponds to x=1 having only one max/min means there can be at most 2 solutions except that u=1 IS a solution, which we can see just from guess & check and since that is the only min/max of f(u), it must be the only solution so we have shown that x=1 is the only solution

[@shahidpervaiz5216]

Y Go that Far, Simple Comparison 👇 x . (e) ^ 1/× = 1 . (e) ^ 1/1 Hence x = 1 👏

[@SyberMath]

Wow! 😜

[@morteza3268]

Xe^(1/x)=e or xe^x=1/e?

[@SyberMath]

I just realized that and fixed it! Sorry

[@morteza3268]

@@SyberMath No problem

[@persassy7076]

Couldn't you see 1 just by looking..?

[@barakathaider6333]

👍

[@MichaelRothwell1]

Nice video but alas the solution is not quite complete since by taking ln of both sides you are restricting x to positive values when the original equation only has the restriction x≠0. This is easily fixed by noting that for x0. My solution was a bit different. xe^(1/x)=e (so x≠0) ⇔e^(1/x)=e/x ⇔eᵗ=et (where t=1/x, t≠0) ⇔eᵗ-et=0 Let f(t)=eᵗ-et Then f(1)=0 [so t=1 is a solution] f'(t)=eᵗ-e f'(t)=0 at t=1 Also f'(t)

[@brahimragueb7149]

Congrats on your 100k followers I just wanna ask you about a problem that I found if you please to help me solving it 1

[@urisinger3412]

There shouldent be a 8 there