What is the answer to solvefor x,xe^y+ye^x=1 ?

The answer to solvefor x,xe^y+ye^x=1 is x=-(\W(y/(e^{-e^{-y)+y}})e^y-1)/(e^y)

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solvefor x,xe^y+ye^x=1

solve for

x, x e^{y} + y e^{x} = 1

x = - \frac{W ( \frac{y}{e ^{- e^{- y} + y}} ) e ^{y} - 1}{e ^{y}}

Solution steps

x e^{y} + y e^{x} = 1

Prepare

x e^{y} + y e^{x} = 1

for Lambert form:

y = (1 - x e^{y}) e^{- x}

Rewrite the equation with

- x + \frac{1}{e ^{y}} = u

and

x = - \frac{e ^{y} u - 1}{e ^{y}}

y = (1 - (- \frac{e ^{y} u - 1}{e ^{y}}) e^{y}) e^{- (- \frac{e ^{y} u - 1}{e ^{y}})}

Simplify

y = e^{\frac{e ^{y} u - 1}{e ^{y}}} e^{y} u

Rewrite

y = e^{\frac{e ^{y} u - 1}{e ^{y}}} e^{y} u

in Lambert form:

e^{u} u = \frac{y}{e ^{- e^{- y} + y}}

Solve

e^{u} u = \frac{y}{e ^{- e^{- y} + y}} : u = W (\frac{y}{e ^{- e^{- y} + y}})

u = W (\frac{y}{e ^{- e^{- y} + y}})

Substitute back

u = - x + \frac{1}{e ^{y}},

solve for

x

Solve

- x + \frac{1}{e ^{y}} = W (\frac{y}{e ^{- e^{- y} + y}}) : x = - \frac{W ( \frac{y}{e ^{- e^{- y} + y}} ) e ^{y} - 1}{e ^{y}}

x = - \frac{W ( \frac{y}{e ^{- e^{- y} + y}} ) e ^{y} - 1}{e ^{y}}

8^{x - 5} = 1024

2 7^{4 x} = 9^{x + 1}

e^{x + 5} = 3

1.1 5^{104} = 1.0 2^{x}

\frac{n + 4}{3 ^{n}} = \frac{2}{3}

What is the answer to solvefor x,xe^y+ye^x=1 ?
The answer to solvefor x,xe^y+ye^x=1 is x=-(\W(y/(e^{-e^{-y)+y}})e^y-1)/(e^y)