What is the answer to solvefor y,y^x=x^y ?

The answer to solvefor y,y^x=x^y is y=-(\W(-(ln(x))/x)x)/(ln(x))

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solvefor y,y^x=x^y

solve for

y, y^{x} = x^{y}

y = - \frac{W ( - \frac{l n ( x )}{x} ) x}{ln ( x )}

Solution steps

y^{x} = x^{y}

Prepare

y^{x} = x^{y}

for Lambert form:

y e^{- \frac{y l n ( x )}{x}} = 1

Rewrite the equation with

- \frac{y ln ( x )}{x} = u

and

y = - \frac{xu}{ln ( x )}

(- \frac{xu}{ln ( x )}) e^{u} = 1

Rewrite

(- \frac{xu}{ln ( x )}) e^{u} = 1

in Lambert form:

e^{u} u = - \frac{ln ( x )}{x}

Solve

e^{u} u = - \frac{ln ( x )}{x} : u = W (- \frac{ln ( x )}{x})

u = W (- \frac{ln ( x )}{x})

Substitute back

u = - \frac{y ln ( x )}{x},

solve for

y

Solve

- \frac{y ln ( x )}{x} = W (- \frac{ln ( x )}{x}) : y = - \frac{W ( - \frac{l n ( x )}{x} ) x}{ln ( x )}

y = - \frac{W ( - \frac{l n ( x )}{x} ) x}{ln ( x )}

9^{x - 2} + 3^{x + 1} = 90

ln (5^{x} - 1) = 3

e^{3 x - 1} = 15

2 5^{2 x + 4} = 62 5^{2 x - 1}

0.8 = \frac{e ^{- 5 + 3 x}}{1 + e ^{- 5 + 3 x}}

What is the answer to solvefor y,y^x=x^y ?
The answer to solvefor y,y^x=x^y is y=-(\W(-(ln(x))/x)x)/(ln(x))