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

The answer to solvefor x,e^{xy}+xy=2 is x=-(\W_{0}(e^2)-2)/y

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

solve for

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

x = - \frac{W _{0} ( e ^{2} ) - 2}{y}

Solution steps

e^{x y} + x y = 2

Prepare

e^{x y} + x y = 2

for Lambert form:

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

Rewrite the equation with

- x y + 2 = u

and

x = - \frac{u - 2}{y}

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

Simplify

1 = e^{u - 2} u

Rewrite

1 = e^{u - 2} u

in Lambert form:

e^{u} u = e^{2}

Solve

e^{u} u = e^{2} : u = W_{0} (e^{2})

u = W_{0} (e^{2})

Substitute back

u = - x y + 2,

solve for

x

Solve

- x y + 2 = W_{0} (e^{2}) : x = - \frac{W _{0} ( e ^{2} ) - 2}{y}

x = - \frac{W _{0} ( e ^{2} ) - 2}{y}

7^{5 x + 7} = 3

1 1^{2 x} = 12 1^{3 x + 1}

12 5^{3 x - 4} = 62 5^{- 3}

5^{2 x + 1} = 7^{x - 3}

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

What is the answer to solvefor x,e^{xy}+xy=2 ?
The answer to solvefor x,e^{xy}+xy=2 is x=-(\W_{0}(e^2)-2)/y