What is the answer to solvefor y,x=(e^y-e^{-y})/2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor y,x=(e^y-e^{-y})/2

solve for

y, x = \frac{e ^{y} - e ^{- y}}{2}

y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

Solution steps

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

Multiply both sides by

2

x \cdot 2 = \frac{e ^{y} - e ^{- y}}{2} \cdot 2

Simplify

x \cdot 2 = e^{y} - e^{- y}

Apply exponent rules

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

Rewrite the equation with

e^{y} = u

x \cdot 2 = u - u^{- 1}

Solve

x \cdot 2 = u - u^{- 1} : u = x + x^{2} + 1, u = x - x^{2} + 1

u = x + x^{2} + 1, u = x - x^{2} + 1

Substitute back

u = e^{y},

solve for

y

Solve

e^{y} = x + x^{2} + 1 : y = ln (x + x^{2} + 1)

Solve

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

y = ln (x + x^{2} + 1), y = ln (x - x^{2} + 1)

5^{x (x - 1)} = 1

e^{3 - 5 x} = 16

y, x = 2^{y}

5^{x + 1} + 5^{x} + 5^{x - 1} = \frac{31}{5}

8^{2 - x} = 2

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