What is the answer to 3x-e^{x-2}-4=0 ?

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

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3x-e^{x-2}-4=0

3 x - e^{x - 2} - 4 = 0

x = - \frac{3 W _{- 1} ( - \frac{1}{3 e ^{\frac{2}{3}}} ) - 4}{3}, x = - \frac{3 W _{0} ( - \frac{1}{3 e ^{\frac{2}{3}}} ) - 4}{3}

Solution steps

3 x - e^{x - 2} - 4 = 0

Prepare

3 x - e^{x - 2} - 4 = 0

for Lambert form:

- 1 = - e^{- x + 2} (3 x - 4)

Rewrite the equation with

- x + \frac{4}{3} = u

and

x = - \frac{3 u - 4}{3}

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

Simplify

- 1 = 3 e^{\frac{3 u - 4}{3} + 2} u

Rewrite

- 1 = 3 e^{\frac{3 u - 4}{3} + 2} u

in Lambert form:

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

Solve

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

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

Substitute back

u = - x + \frac{4}{3},

solve for

x

Solve

- x + \frac{4}{3} = W_{- 1} (- \frac{1}{3 e ^{\frac{2}{3}}}) : x = - \frac{3 W _{- 1} ( - \frac{1}{3 e ^{\frac{2}{3}}} ) - 4}{3}

Solve

- x + \frac{4}{3} = W_{0} (- \frac{1}{3 e ^{\frac{2}{3}}}) : x = - \frac{3 W _{0} ( - \frac{1}{3 e ^{\frac{2}{3}}} ) - 4}{3}

x = - \frac{3 W _{- 1} ( - \frac{1}{3 e ^{\frac{2}{3}}} ) - 4}{3}, x = - \frac{3 W _{0} ( - \frac{1}{3 e ^{\frac{2}{3}}} ) - 4}{3}

(3^{x + 1}) (4^{x + 2}) = 5^{2 x - 1}

2 e^{x} = 7 e^{x} - 3

\frac{1}{64} = 2^{x}

12 5^{x - 1} = 2 5^{x}

x, x y e^{\frac{y}{x}} = 5

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