What is the answer to solvefor x,x+y+e^{xy}=0 ?

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

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solvefor x,x+y+e^{xy}=0

solve for

x, x + y + e^{x y} = 0

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

Solution steps

x + y + e^{x y} = 0

Prepare

x + y + e^{x y} = 0

for Lambert form:

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

Rewrite the equation with

- x y - y^{2} = u

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

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

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

Substitute back

u = - x y - y^{2},

solve for

x

Solve

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

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

2^{2 x + 1} = 3 (2^{x}) + 2

3^{x} + 16 = 9 \cdot 3^{x}

4^{v + 9} + 5 = 38

- 10 \cdot 5^{8 - 4 x} - 7 = - 12

3^{x - 1} + 3^{x + 1} = 90

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