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

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

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solvefor y,x^y=y^x

solve for

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

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

Solution steps

x^{y} = y^{x}

Prepare

x^{y} = y^{x}

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Rewrite

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

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 )}

k, - 2 e^{- 4 k} = 1

2^{x + 5} = 2^{2 x + 1}

5^{x + 1} - 5^{x - 1} = 24

e^{(- 2 x)} - 2 e^{- x} = 0.5

x, 2 = e^{x}

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