A Nice Homemade Exponential Equation

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[chaosredefined]

When we get t^2 + ln(t) = 1...
We want to get this in the form of thing * eexp(thing) = constant.
t^2 + ln(t) = 1
2t^2 + ln(t^2) = 2
exp(2t^2) * t^2 = exp(2)
2t^2 * exp(2t^2) = 2*exp(2)
Applying the Lambert-W function to both sides
2t^2 = 2
t^2 = 1
t = 1 or t = -1. As t was in a logarithm, t =/= -1. Hence t = 1.
We defined t as ln(x), so ln(x) = 1, or x = e.

No need to guess.

[BillKilmerslayer]

Why not raise both sides to the power of e? It simplifies to x^x^x=e^e^e, ipso facto x=e

[kuriana]

Excellent. I really like the way you present your solution. Cheers.

[rrivierareject]

I always appreciate seeing all of the steps. It's helpful for convincing that we've found every possible solution.

[Wangkheimayum]

hit and trial method works x=e sometimes you dont have to think too deep for such problem

[Paul-]

There’s also a complex solution at x = e^(i[pi]+1). That’s just the principal branch, of course.

[tetramur]

ln x = t, x = e^t
(e^t)^(e^t)^t = e^e
(e^t)^(e^t^2) = e^e
e^(t*e^t^2) = e^e, or:
t*e^t^2 = e (1)
2t^2*e^(2t^2) = 2e^2
2t^2 = W(2e^2) = 2, or t = ±1
back-substitution reveals that t = -1 is not a root of (1). Therefore t = 1, or x = e^t = e

[mcbeaulieu]

At first glance, x = e

x^ln(x).ln(x) = e
Trial and error :
e^ln(e).ln(e) = (e^1).1 = e

Now watching for the proper technique 😅

[giuseppemalaguti]

Pongo lnx=t... t^2+lnt=1...una soluzione è t=1(x=e), puo esserci un'altra soluzione per 0<t<1....in generale x=e^(W(2e^2)/2)

[tetsuyaikeda]

increasing function and decreasing function are going to cross at only single point on XY space.

[arbenkellici]

Helllo profesor! I have been watching your videos and you are doing an excellent job As a matter of fact, I need your help with a math exercise Let a be a positive real number We have to prove that a +1/a>=2 Please give me a solution as soon as possible
Thank you!

[nasrullahhusnan]

t²+lnt=1
By inspection it is clear that the solution is t=1 as ln1=0.
Is there any general method to solve if the solution is not so clear?

[zawatsky]

Число экспонируется само на себя, значит x=x^ln(x). Но x=x¹ => ln(x)=1. Log(e;x)=1 => e¹=x. Отсюда, x=e, что можно посчитать и по основанию. Но e¹≠e^e, поскольку e≠1. Вывод: решений нет.

[prof.mohamad]

Si on remplace des le debut e par e^loge le probleme devient tres facile et evident et par comparaison on voit que x=e

[prof.mohamad]

On peut mettre des le debut e = e^loge car loge=1 alors on peut voir tout de suite que x=e

[andywalls]

Also, you could have use the W-Lambert function, of the next way:
x^(x^lnx) = e^e
ln(x^(x^lnx)) = ln(e^e)
(x^lnx)*lnx = e
lnx * x^lnx = 1e^1
W(lnx * x^lnx) = W(1e^1)
ln x = 1
e^lnx = e^1
x = e

[prof.mohamad]

X^x^logx = e^e^loge
Alors x=e
C est evident