What is the answer to solvefor y,x=(1+e^y)/(1-e^y) ?

The answer to solvefor y,x=(1+e^y)/(1-e^y) is y=ln(-(1-x)/(1+x)) {x 1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor y,x=(1+e^y)/(1-e^y)

solve for

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

y = ln (- \frac{1 - x}{1 + x}) {x < - 1 or x > 1}

Solution steps

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

Multiply both sides by

1 - e^{y}

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

Simplify

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

Expand

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

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

Move

x

to the right side

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

Move

e^{y}

to the left side

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

Factor

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

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

Divide both sides by

- (x + 1)

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

Apply exponent rules

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

Verify Solutions:

y = ln (- \frac{1 - x}{1 + x}) {x < - 1 or x > 1}

The solution is

y = ln (- \frac{1 - x}{1 + x}) {x < - 1 or x > 1}

5 (2)^{8 t} = 15

9 e^{x} = 1

x, 4^{x + 1} + 4^{x} = 160

2^{4} = 8^{x}

\frac{8 1 ^{3 x + 2}}{24 3 ^{- x}} = 3^{4}

What is the answer to solvefor y,x=(1+e^y)/(1-e^y) ?
The answer to solvefor y,x=(1+e^y)/(1-e^y) is y=ln(-(1-x)/(1+x)) {x<-1\lor x>1}