What is the answer to e^{-6x}=1-4x ?

The answer to e^{-6x}=1-4x is x=(2\W_{-1}(-3/(2e^{3/2)})+3)/(12),x=(2\W_{0}(-3/(2e^{3/2)})+3)/(12)

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e^{-6x}=1-4x

e^{- 6 x} = 1 - 4 x

x = \frac{2 W _{- 1} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}, x = \frac{2 W _{0} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

Decimal Notation

x = 0, x = 0.14570 \dots

Solution steps

e^{- 6 x} = 1 - 4 x

Prepare

e^{- 6 x} = 1 - 4 x

for Lambert form:

1 = (1 - 4 x) e^{6 x}

Rewrite the equation with

6 x - \frac{3}{2} = u

and

x = \frac{2 u + 3}{12}

1 = (1 - 4 (\frac{2 u + 3}{12})) e^{6 (\frac{2 u + 3}{12})}

Simplify

1 = e^{\frac{2 u + 3}{2}} (1 - \frac{2 u + 3}{3})

Rewrite

1 = e^{\frac{2 u + 3}{2}} (1 - \frac{2 u + 3}{3})

in Lambert form:

e^{u} u = - \frac{3}{2 e ^{\frac{3}{2}}}

Solve

e^{u} u = - \frac{3}{2 e ^{\frac{3}{2}}} : u = W_{- 1} (- \frac{3}{2 e ^{\frac{3}{2}}}), u = W_{0} (- \frac{3}{2 e ^{\frac{3}{2}}})

u = W_{- 1} (- \frac{3}{2 e ^{\frac{3}{2}}}), u = W_{0} (- \frac{3}{2 e ^{\frac{3}{2}}})

Substitute back

u = 6 x - \frac{3}{2},

solve for

x

Solve

6 x - \frac{3}{2} = W_{- 1} (- \frac{3}{2 e ^{\frac{3}{2}}}) : x = \frac{2 W _{- 1} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

Solve

6 x - \frac{3}{2} = W_{0} (- \frac{3}{2 e ^{\frac{3}{2}}}) : x = \frac{2 W _{0} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

x = \frac{2 W _{- 1} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}, x = \frac{2 W _{0} ( - \frac{3}{2 e ^{\frac{3}{2}}} ) + 3}{12}

z, c = e^{2 x + 2 y + z}

3^{x + 2} - 3^{x + 1} + 3^{x} = 63

3^{x + 4} = 9

5^{2 x - 1} = \frac{1}{125}

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

What is the answer to e^{-6x}=1-4x ?
The answer to e^{-6x}=1-4x is x=(2\W_{-1}(-3/(2e^{3/2)})+3)/(12),x=(2\W_{0}(-3/(2e^{3/2)})+3)/(12)