Another Nice Exponential Equation

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[SweetSorrow]

This has Lambert W function written all over it.😊

[mr.d]

*This can be solved even faster by using the Lambert W function:*
*x • e^(1/x) = e /^(-1)*
*(1/x) • e^(-1/x) = e^(-1) /•(-1)*
*(-1/x) • e^(-1/x) = (-1) • e^(-1) /W*
*-1/x = -1 /÷(-1)*
*1/x = 1 /^(-1)*
*x = 1.*
*We can know that this is the only one real solution because -1 is the only real value of W(-1/e). In fact, -1/e is the smallest real number with a real Lambert W function, there are two real values for W(x) when x is between -1/e and 0.*

[pwmiles]

x e^(1/x) = e

You can make the substitution
y = 1/x
(1/y) e^y = e
y e^(-y) = 1/e
-y e^(-y) = -1/e
-y = W(-1/e)

-1/e is the minimum value of x e^x, attained at x = -1. Because d/dx(x e^x) = (1+x)e^x = 0 at x=-1.
Hence -y=-1, y=1, x = 1 is the only real solution.

[renesperb]

It took me a few seconds to see that of course x= 1 is a solution.
If you look at the graphs it is also evident that it is the only real solution.

[freddyalvaradamaranon]

Interesante video de una ecuacion con logaritmos naturales, muchas gracias por compartir 😊❤😊.

[SidneiMV]

(1/x)e^(-1/x) = e⁻¹
(-1/x)e^(-1/x) = -1e⁻¹
W[(-1/x)e^(-1/x)] = W(-1e⁻¹)

-1/x = -1 => *x = 1*

*x = -1/W(-1e⁻¹)*

a possible value for W(-1e⁻¹) is -1
so x = -1/(-1) => *x = 1*

[ChristopherBitti]

I don't know anything about this Lambert function. What I know is that the derivative of f(x) = xe^(1/x) is [(x-1)e^(1/x)]/x

When x < 0 the derivative is positive so f is increasing
f is not defined at 0
When 0 < x < 1 the derivative is negative so f is decreasing
When x = 1 the derivative is 0
When x > 1 the derivative is positive so f is increasing

Thus, when x = 1, f has a local min and f(1) is the minimum of f on the interval (0, infinity).

Since f(1) = e, 1 is the only solution on the interval (0, infinity).

On the interval x < 0, there are clearly no solutions because f(x) is negative.


Thus, x = 1 is the only solution.

[beniocabeleleiraleila]

xe^(1/x) = e

1/x * e^(-1/x) = e^-1

-1/x * e^(-1/x) = -e^-1

using the lambert W function

-1/x = -1
x=1

[henricovsky]

At 7:20, if x<0 then f’(x) is positive again. Unless you’re still assuming x>0

[danielputrahandana]

the most logical is to put 1 since 1(e)^1/1= e, unless u wanna go around the world to proof it.good video tho, it teach us the lambert wfunction

[humfff]

Why not just say "By observation"? But beware, as often simple looking equations have multiple complex roots.

[andretewem]

No need of Lambert fontion for resolution (too "hard")

[Radii_]

If we multiply the power of LHS and RHS by x then we get xe=e^x from which we get x=1

[official_mosfet]

People breaking their heads making procedures just to realize x is 1. Always guess first cause i didnt have to do any of this man💀💀💀

[someperson]

Clearly if x is a solution, then x > 0. Let f(x) =: ln(x) + 1/x, for x > 0. As you show, the problem is equivalent to solving f(x) = 1. Now, f'(x) = (x -1)/x^2. So, f(x) is strictly decreasing for x in (0, 1] and strictly increasing for x in [1, infinity). Thus, f(x) has an absolute minimum at x = 1 and only at x =1. Since f(1) = 1, the only x where f(x) = 1 is
x = 1.

[hzhz]

You already lost me in the first step where you say "...and obviously this equals to 1"

[krishnarao]

x = e^x-1/x

lnx = x-1/x

ln x^x = x-1

e^x-1 = x^x

e^x = ex^x

(e/x)^x = e

so only possible is x = 1 :D

[RealQinnMalloryu]

w^{xe*1/x}^w=w^{xe ➖ 1xe+1/xe ➖ 1xe+1}^w=w^{xe ➖ 1/xe+1}^w (w^{xe ➖ 1/xe+1}^w}) .