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

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

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solvefor y,2x=e^y-e^{-y}

solve for

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

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

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

e^{y} = u

2 x = u - u^{- 1}

Solve

2 x = 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)

3125 = 2^{x}

3^{5 x - 1} = 8

2^{2 + y} - 2^{1 + y} = 16

66.1 = 120 - 120 e^{k (0.5)}

0 = 0.9 8^{X}

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