What is the answer to e^{-x}+2x-1=0 ?

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

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e^{-x}+2x-1=0

e^{- x} + 2 x - 1 = 0

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

Decimal Notation

x = - 1.25643 \dots, x = 0

Solution steps

e^{- x} + 2 x - 1 = 0

Prepare

e^{- x} + 2 x - 1 = 0

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

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

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

Substitute back

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

solve for

x

Solve

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

Solve

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

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

2^{8^{x - 2}} = 4^{4^{x - 1}}

4^{2 x + 1} - 65 (4^{x}) + 16 = 0

e^{(9 x)} - 5 e^{(3 x)} + 4 = 0

2^{3 x - 8} = 16

x^{(\frac{0.72}{( x - 1 )})} = 2

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