Solving An Exponential Equation With A Parameter

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[@pageboysam]

8:33 1 < a < e^(4/e^2) has 3 solutions! The third solution is very far out to the right.

[@SyberMath]

at infinity? 😜

[@pageboysam]

@@SyberMath Check it!
f(x) = x^((ln x) / x)
f(0.82) ~= f(1.27) ~= f(1000) ~= 1.048

[@yuryp6975]

​@@SyberMath One solutions is x somewhere between 0 and 1, and 2 solutions where x is between 1 and infinity, since the function increases and then decreased and approaches 1

[@yuichiro12]

Yes, similarly, a = e^(4/e^2) has 2 solutions and a > e^(4/e^2) has one solution.

[@Don-Ensley]

x₃=e^(-2 W₋₁(-½√[ln(a)]))

[@Don-Ensley]

One point for
a > e^[4 /(e²)]:
x₁ =e^(-2 W₀(½√[ln(a)]))
A family of values.
Another one point solution
a=1:
x = 1
A single point.
Two points for
a = e^[4 /(e²)]:
x₁ =e^(-2 W₀(1/e])
x₂ =e²
The only two point solution.
Three points for
1 < a < e^[4 /(e²)]:
x₁,₂=e^(-2 W₀(± ½√[ln(a)]))
x₃=e^(-2 W₋₁(-½√[ln(a)]))
A family of values.

[@RedRad1990]

Since this function approaches _+∞_ as _x_ approaches zero, I'm pretty sure, you get two solutions for _a = e ^ [ 4/( e^2 ) ]_ (one is _e^2,_ the other one is approximately 0.57)
Similarly, for all _a > e ^ [ 4/( e^2 ) ]_ there exists exactly one real solution that is on the interval _(0, 0.57...)_

[@scottleung9587]

Cool!

[@rakenzarnsworld2]

x = e

[@giuseppemalaguti435]

X=e^(-2W((-e^k)/2))..k=(lnlna)/2

[@thexavier666]

Dark background always please

[@SidneiMV]

(1/x)(lnx)lnx = lna
(1/√x)lnx = √lna
x^(-1/2)lnx = √lna
x^(-1/2)lnx^(-1/2) = -1/2√lna
lnx^(-1/2) = W(-1/2√lna)
x^(-1/2) = e^W(-1/2√lna)
x = e^[(-2)W(-1/2√lna)]

[@MAZHOR1113]

ура я снова первый🎉🎉🎉