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

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

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e^x=2x+1

e^{x} = 2 x + 1

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

Decimal Notation

x = 1.25643 \dots, x = 0

Solution steps

e^{x} = 2 x + 1

Prepare

e^{x} = 2 x + 1

for Lambert form:

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

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})}

Simplify

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

Rewrite

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

Solve

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

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

e^{x} = 2 x + 2

2^{3 x - 4} = 18

3 \cdot (2)^{x - 1} - 2 = 0

6^{2 x} + 6^{x} - 6 = 0

9 (3^{2 x}) - 22 (3^{x}) - 15 = 0

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