What is the answer to solvefor y,xe^y+ye^x=3 ?

The answer to solvefor y,xe^y+ye^x=3 is y=-(\W(x/(e^{-3e^{-x)+x}})e^x-3)/(e^x)

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solvefor y,xe^y+ye^x=3

solve for

y, x e^{y} + y e^{x} = 3

y = - \frac{W ( \frac{x}{e ^{- 3 e^{- x} + x}} ) e ^{x} - 3}{e ^{x}}

Solution steps

x e^{y} + y e^{x} = 3

Prepare

x e^{y} + y e^{x} = 3

for Lambert form:

x = (3 - y e^{x}) e^{- y}

Rewrite the equation with

- y + \frac{3}{e ^{x}} = u

and

y = - \frac{e ^{x} u - 3}{e ^{x}}

x = (3 - (- \frac{e ^{x} u - 3}{e ^{x}}) e^{x}) e^{- (- \frac{e ^{x} u - 3}{e ^{x}})}

Simplify

x = e^{\frac{e ^{x} u - 3}{e ^{x}}} e^{x} u

Rewrite

x = e^{\frac{e ^{x} u - 3}{e ^{x}}} e^{x} u

in Lambert form:

e^{u} u = \frac{x}{e ^{- 3 e^{- x} + x}}

Solve

e^{u} u = \frac{x}{e ^{- 3 e^{- x} + x}} : u = W (\frac{x}{e ^{- 3 e^{- x} + x}})

u = W (\frac{x}{e ^{- 3 e^{- x} + x}})

Substitute back

u = - y + \frac{3}{e ^{x}},

solve for

y

Solve

- y + \frac{3}{e ^{x}} = W (\frac{x}{e ^{- 3 e^{- x} + x}}) : y = - \frac{W ( \frac{x}{e ^{- 3 e^{- x} + x}} ) e ^{x} - 3}{e ^{x}}

y = - \frac{W ( \frac{x}{e ^{- 3 e^{- x} + x}} ) e ^{x} - 3}{e ^{x}}

4^{x + 1} = 8^{2 x + 3}

(1000) 1.0 5^{3 x} = 3000

4 9^{x} = 21

320000 = 250342 e^{0.012 x}

3^{5 x + 34} = 81

What is the answer to solvefor y,xe^y+ye^x=3 ?
The answer to solvefor y,xe^y+ye^x=3 is y=-(\W(x/(e^{-3e^{-x)+x}})e^x-3)/(e^x)