Can We Solve A Beautiful Equation 😊

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[@SidneiMV]

x^x = 2^(1/x)
x^x² = 2
x²^x² = 2²
x² = 2 => *x = √2*
(x = -√2 is not valid)

[@xualain3129]

Bravo! I got exactly the same idea as yours.

[@mcwulf25]

This is how I solved it 👍

[@Anmol_Sinha]

why is x = -√2 not valid?

[@mcwulf25]

@Anmol_Sinha when both sides were squared, that extraneous solution was added. The rhs is always positive. The lhs must also therefore be positive.

[@SidneiMV]

@@mcwulf25 exactly. tks

[@allanmarder456]

This can also be solved using the Lambert W function. First take ln of both sides. Thus x*ln(x)=(1/x)*ln(2) or (x*2)*ln(x) = ln(2). Now let t=x^2.
That gives t*ln(t^(1/2)) = ln(2), Simplifying gives t*ln(t) = 2*ln(2). This equation can be written as ln(t)* e^ln(t) = (ln(2))*e^ln(2). If we take
the lambert W of both sides W( ln(t)* e^ln(t)) = W( (ln(2))*e^ln(2)) or ln(t)= ln(2) or t=2 or x^2 = 2 or x=sqrt(2). The advantage of this
method is the argument of W( (ln(2))*e^ln(2)) is in the branch of the W function where there is only a single answer. Thus x=sqrt(2)
is the only answer and the calculus approach can be avoided.

[@YouTube_username_not_found]

Actually, you need calculus to find out how many solutions in each branch of W function.

[@HarisRehmanGG]

You didn't tell us the bprp fish introduction before using it 😭

[@Viki13]

​@TH-cam_username_not_found you can just use the definition of the W function

[@YouTube_username_not_found]

@@Viki13 Not sure how one can do that.

[@Viki13]

@@TH-cam_username_not_found W(2ln(2))=W(e^(ln(2))•ln(2))=ln(2) by definition since W(xe^x)=x

[@rakeshsrivastava1122]

Hope your pet cat is doing fine.

[@lagomoof]

Made me wonder if there's a straight line that crosses x^x^2 in three places and yes, there are infinitely many.
One simple example is: y = -x/3 + 1.01
If 0^0^2 could be said to be exactly 1, and that was then allowed to count as a crossing point, then the 1.01 could just be 1, but that would only anger the mathematical gods. And many fellow TH-cam commenters.
As for finding those crossing points algebraically, I don't think even WolframAlpha can help here. Except if it does so numerically.

[@nasrullahhusnan2289]

x^x=2^(1/x)
Raise to x --> x^x²=2
y=x² --> x=y^(½)
[y^(½)]^y=2
y^(½y)=2
½y^(½y)=1 --> ½y=1
y=2
x²=2 --> x=±sqrt(2)

[@walterwen2975]

Solve A Beautiful Equation: xˣ = 2¹⸍ˣ; x = ?
x ≠ 0; (xˣ)ˣ = (2¹⸍ˣ)ˣ, x^x² = 2, (x^x²)² = 2², (x²)^x² = 2², x² = 2; x = ± √2
Answer check:
2¹⸍ˣ = 2^(± 1/√2) = [(√2)²]^(± 1/√2) = (√2)^(2/√2) = (√2)^(√2) = xˣ; Confirmed
Final answer:
x = √2 or x = - √2

[@KevinKevin-dh3wk]

Is x=2 correct??

[@Viki13]

No

[@phill3986]

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