What is the answer to solvefor x,e^{xy}+y=x-1 ?

The answer to solvefor x,e^{xy}+y=x-1 is x=-(\W(-y/(e^{-y^2-y)})+y(-1-y))/y

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solvefor x,e^{xy}+y=x-1

solve for

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

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

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

- x y - y (- 1 - y) = u

and

x = - \frac{u + y ( - 1 - y )}{y}

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

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

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

Substitute back

u = - x y - y (- 1 - y),

solve for

x

Solve

- x y - y (- 1 - y) = W (- \frac{y}{e ^{- y^{2} - y}}) : x = - \frac{W ( - \frac{y}{e ^{- y^{2} - y}} ) + y ( - 1 - y )}{y}

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

2 e^{2 x} - e^{- x} = 0

4^{x} - 2^{x + 1} = 48

(7^{- x})^{2} = 343

5^{x + 2} = 4^{x}

9^{x} = 3

What is the answer to solvefor x,e^{xy}+y=x-1 ?
The answer to solvefor x,e^{xy}+y=x-1 is x=-(\W(-y/(e^{-y^2-y)})+y(-1-y))/y