What is the answer to e^x+e^{-x}=2e ?

The answer to e^x+e^{-x}=2e is x=ln((2e+sqrt(4e^2-4))/2),x=ln((2e-sqrt(4e^2-4))/2)

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

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

x = ln (\frac{2 e + 4 e ^{2} - 4}{2}), x = ln (\frac{2 e - 4 e ^{2} - 4}{2})

Decimal Notation

x = 1.65745 \dots, x = - 1.65745 \dots

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

e^{x} = u

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

Solve

u + u^{- 1} = 2 e : u = \frac{2 e + 4 e ^{2} - 4}{2}, u = \frac{2 e - 4 e ^{2} - 4}{2}

u = \frac{2 e + 4 e ^{2} - 4}{2}, u = \frac{2 e - 4 e ^{2} - 4}{2}

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = \frac{2 e + 4 e ^{2} - 4}{2} : x = ln (\frac{2 e + 4 e ^{2} - 4}{2})

Solve

e^{x} = \frac{2 e - 4 e ^{2} - 4}{2} : x = ln (\frac{2 e - 4 e ^{2} - 4}{2})

x = ln (\frac{2 e + 4 e ^{2} - 4}{2}), x = ln (\frac{2 e - 4 e ^{2} - 4}{2})

3^{(6 + 3 x)} = \frac{1}{27}

2 0^{x} = 4

5000 = 3^{n - 1}

1 6^{2 x - 1} = 6 4^{x + 1}

What is the answer to e^x+e^{-x}=2e ?
The answer to e^x+e^{-x}=2e is x=ln((2e+sqrt(4e^2-4))/2),x=ln((2e-sqrt(4e^2-4))/2)