A Nice Homemade Exponential Equation | x^x=y^{1/y}

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[toma]

Why y=x^t substitution? Maybe y=x^2 +x^t or any other what comes to my mind?

[Paul-]

I like the y = x^t trick. I attempted the problem before watching, and it was easy enough to see that x = y = +/- 1 is a solution. (I don’t see why you discarded -1 as a solution, since it satisfies the original equation). I didn’t know where to take it from there, so your trick was helpful.

[mathsandsciencechannel]

It always important to check the domain of an equation before attempting it. Nice solution sir

[laurentreouven]

The last question is the most interesting one 🙂

[musicsubicandcebu]

Interesting similarity between the current parameterization and that of x^y = y^x

[chaosredefined]

I think t = xy makes a much cleaner substitution.

x^x = y ^(1/y)
x^xy = y

Let t = xy. Therefore, y = t/x
x^t = t/x
x^(t+1) = t
x = t^(1/(t+1))

xy = y t^(1/(t+1))
t = y t^(1/(t+1))
t * t^(-1/(t+1)) = y
t^(t/(t+1)) = y

Solutions are of the form (x, y) = (t^(1/(t+1)), t^(t/(t+1)))

To find dy/dx, Id say implicit differentiation is, once again, cleaner.
x^x = y^(1/y)
x ln x = (ln y)/y
d/dx (x ln x) = ln x + 1 (By product rule)
d/dx ((ln y)/y) = y' (1 - ln y)/y^2 (By chain rule, and quotient rule)
ln x + 1 = y' (1 - ln y)/y^2
y^2 (ln x + 1)/(1 - ln y) = y'

[neuralwarp]

Ah, so you could work back from that solution and find an expression for W(t) as a function of t.