What is the answer to -e^{x+1}=2x+1 ?

The answer to -e^{x+1}=2x+1 is x=-(2\W_{0}((e^{1/2})/2)+1)/(2)

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

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

x = - \frac{2 W _{0} ( \frac{e ^{\frac{1}{2}}}{2} ) + 1}{2}

Decimal Notation

x = - 1

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

- x - \frac{1}{2} = u

and

x = - \frac{2 u + 1}{2}

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = \frac{e ^{\frac{1}{2}}}{2}

Solve

e^{u} u = \frac{e ^{\frac{1}{2}}}{2} : u = W_{0} (\frac{e ^{\frac{1}{2}}}{2})

u = W_{0} (\frac{e ^{\frac{1}{2}}}{2})

Substitute back

u = - x - \frac{1}{2},

solve for

x

Solve

- x - \frac{1}{2} = W_{0} (\frac{e ^{\frac{1}{2}}}{2}) : x = - \frac{2 W _{0} ( \frac{e ^{\frac{1}{2}}}{2} ) + 1}{2}

x = - \frac{2 W _{0} ( \frac{e ^{\frac{1}{2}}}{2} ) + 1}{2}

5^{x} = 5^{2 x + 1}

1 0^{- 2 x} = \frac{1}{10000}

5^{x^{2}} = (\frac{1}{25})^{x + 3}

\frac{2 ^{x}}{3 ^{x + 1}} = 1

4^{(2 x - 1)} = \frac{1}{2}

What is the answer to -e^{x+1}=2x+1 ?
The answer to -e^{x+1}=2x+1 is x=-(2\W_{0}((e^{1/2})/2)+1)/(2)