What is the answer to solvefor x,e^{2xy}+2y=2x+2 ?

The answer to solvefor x,e^{2xy}+2y=2x+2 is x=-(\W(-y/(e^{-2y^2+2y)})+y(2-2y))/(2y)

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solvefor x,e^{2xy}+2y=2x+2

solve for

x, e^{2 x y} + 2 y = 2 x + 2

x = - \frac{W ( - \frac{y}{e ^{- 2 y^{2} + 2 y}} ) + y ( 2 - 2 y )}{2 y}

Solution steps

e^{2 x y} + 2 y = 2 x + 2

Prepare

e^{2 x y} + 2 y = 2 x + 2

for Lambert form:

1 = (2 x + 2 - 2 y) e^{- 2 x y}

Rewrite the equation with

- 2 x y - y (2 - 2 y) = u

and

x = - \frac{u + y ( 2 - 2 y )}{2 y}

1 = (2 (- \frac{u + y ( 2 - 2 y )}{2 y}) + 2 - 2 y) e^{- 2 (- \frac{u + y ( 2 - 2 y )}{2 y}) y}

Simplify

1 = e^{u + y (2 - 2 y)} (- \frac{u + y ( 2 - 2 y )}{y} + 2 - 2 y)

Rewrite

1 = e^{u + y (2 - 2 y)} (- \frac{u + y ( 2 - 2 y )}{y} + 2 - 2 y)

in Lambert form:

e^{u} u = - \frac{y}{e ^{- 2 y^{2} + 2 y}}

Solve

e^{u} u = - \frac{y}{e ^{- 2 y^{2} + 2 y}} : u = W (- \frac{y}{e ^{- 2 y^{2} + 2 y}})

u = W (- \frac{y}{e ^{- 2 y^{2} + 2 y}})

Substitute back

u = - 2 x y - y (2 - 2 y),

solve for

x

Solve

- 2 x y - y (2 - 2 y) = W (- \frac{y}{e ^{- 2 y^{2} + 2 y}}) : x = - \frac{W ( - \frac{y}{e ^{- 2 y^{2} + 2 y}} ) + y ( 2 - 2 y )}{2 y}

x = - \frac{W ( - \frac{y}{e ^{- 2 y^{2} + 2 y}} ) + y ( 2 - 2 y )}{2 y}

e^{2 x} - 2 e^{x} = 15

8^{x} = 0.25

2^{x - 6} = 18

12 \cdot 3^{- x} = \frac{4}{3}

2^{x + 1} + 2^{x + 2} = 96

What is the answer to solvefor x,e^{2xy}+2y=2x+2 ?
The answer to solvefor x,e^{2xy}+2y=2x+2 is x=-(\W(-y/(e^{-2y^2+2y)})+y(2-2y))/(2y)