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

The answer to solvefor x,xy-e^x+e^y=0 is x=-(\W(-1/(e^{\frac{e^y){y}}y})y+e^y)/y

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,xy-e^x+e^y=0

solve for

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

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

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = - \frac{1}{e ^{\frac{e ^{y}}{y}} y}

Solve

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

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

Substitute back

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

solve for

x

Solve

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

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

x, 5 (2 5^{x}) = 3.12 5^{y}

5^{x + 3} = 7^{4 - 3 x}

2 7^{x - 1} = 9

3^{2 x - 3} \cdot 9 = 1

12 = 21 e^{20 x}

What is the answer to solvefor x,xy-e^x+e^y=0 ?
The answer to solvefor x,xy-e^x+e^y=0 is x=-(\W(-1/(e^{\frac{e^y){y}}y})y+e^y)/y