You are a true maths genius if you can solve this...

[@brnmcc01]

You can also simplify the answer a little bit to take the natural log of the square root of 3, since a^(log(x)/2) is the same as a^(log(sqrt(x))). So X would be equal to e^(W(log(sqrt(3)))×2) in plain text.
2.1247993830225090335085882479846816946346241964809854993870214081

[@marilynman]

You have a typo on: a^(x/2) is the same as a^(sqrt(x))
It should be: a^(log(x)/2) is the same as a^(log(sqrt(x)))

[@brnmcc01]

@@marilynman You're right, fixed it, thx.

[@rishabvaid2059]

1. Iterative Approximation:We can rewrite the equation as:√x = logx(3)or equivalently:x = (logx(3))²We can start with an initial guess for x (let's say x = 2) and iteratively refine it using the equation above.Iteration 1: x = (log₂(3))² ≈ 2.46Iteration 2: x = (log2.46(3))² ≈ 2.2Iteration 3: x = (log2.2(3))² ≈ 2.3If we keep iterating, the value will converge towards a solution. This method is not very efficient, but it demonstrates a numerical approach. A more sophisticated numerical method like the Newton-Raphson method would converge much faster.
While there isn't a neat, closed-form solution, numerical methods provide approximations. Using iterative techniques or software capable of handling the Lambert W function will yield a solution close to x ≈ 2.2. A more advanced numerical approach would give you a more precise answer.

[@RyanLewis-Johnson-wq6xs]

X^Sqrt[X]=3 X=e^(2W(Ln(3)/2))=e^(2W(Log[e^2,3]))

[@ganymed1236]

Well done! 😊

[@RyanLewis-Johnson-wq6xs]

Ln(3)/2=Log[e^2,3] Input I just proved that it could be simplified.
Ln(3)/2 =Log(e^2, 3)
Results
True
True

[@Fc.roblox_1152]

from scipy.optimize import fsolve
# Define the equation y^(2y) = 3
def equation(y):
return y**(2*y) - 3
# Find the value of y that solves the equation
y_solution = fsolve(equation, 1.2)[0] # Initial guess is 1.2
# Calculate x from y
x_solution = y_solution**2
y_solution, x_solution

[@berntfritiofulveborn3163]

Really nice presentation but could you link to some good Lambert function calculators 👍🙂

[@rexford9019]

Or you can raise both sides to the 1/√x power and then multiply both sides by -1/2, then apply W function.

[@_Grative]

This is just one of the solutions, when you take the natural log you need to add that 2niπ brother

[@ManojkantSamal]

*=read as square root
^= read as to the power
According to the question
X^(*x)=3Take the square
(X^(*x)}^2=3^2
X^x=3^2
(X^x)^(1/x)=(3^2)^(1/x)
X=3^(2/x)
Take the log
X=9^(1/x)

[@liamipiami1468]

You need to have x on one side you cant just say x = x

[@PerezKwadwoAwuah]

I don’t know if my method is also right but I got a different answer though. Let’s see
Let y = sqrt(x)
yln(x) = ln(3)
(1/2)yln(x) = (1/2)ln(3)
yln(y) = ln[sqrt(3)]
e^(yln(y)) = e^(ln[sqrt(3)])
ye^y = sqrt(3)
W(ye^y) = W(sqrt(3))
y = W(sqrt(3))
But y = sqrt(x)
sqrt(x) = W(sqrt(3))
(sqrt(x))^2 = [ W(sqrt(3)) ]^2
x = [ W(sqrt(3)) ]^2
Is it?😶

[@Carlos76672]

Yes it is

[@SidneiMV]

√x^√x = √3
√xln√x = ln√3
ln√x = W(ln√3)
√x = e^W(ln√3)
x = e^[2W(ln√3)]

[@porta_patrols]

cool!

[@RyanLewis-Johnson-wq6xs]

It’s in my head

[@RealQinnMalloryu4]

(x ➖ 3x+3).

[@lucatherine4089]

(ln √x) e^(ln √x) = ½ ln 3
⇒ ln √x = W(½ ln 3) = (½ ln 3)/√x
⇒ x = [(½ ln 3)/ W(½ ln 3)]²,
∴ x = [(½ ln 3)/ W_0(½ ln 3)]² ≈ 2.1247993830225090335085882479846816+

[@ManojkantSamal]

X is in between 2 &3
Respected Sir, Good evening... Pls get me the whole of Lambert aw function....

[@nishasunil207]

How you got that approximate answer?

[@luizfabiano4334]

computer

[@liejdbdj-g6]

미국 어학연수 생활비 대출 좀 해줘 대출좀 그런데 난 대출 안좋아하고 소출 좋아해 소출

[@ЛеонидФедяков-ъ9я]

Too many words.

[@liejdbdj-g6]

랩 좀 그만해 😠

[@prollysine]

let u=Vx , Vx*lnx=ln3 , u*ln(u^2)=ln3 , 2*u*lnu=ln3 , u*lnu=(ln3)/2 , lnu=W((ln3)/2) , u=e(W((ln3)/2) , Vx=e^(0.376838695) , Vx=~ 1.45767 ,
Vx=~ 1.45767^2 , x=`~ 2.1248 , test , x^Vx= 2.1248^1.45767 --> 3 , OK ,