What is the answer to roots e^x-3x-7 ?

The answer to roots e^x-3x-7 is x=-(3\W_{-1}(-1/(3e^{7/3)})+7)/3 ,x=-(3\W_{0}(-1/(3e^{7/3)})+7)/3

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roots e^x-3x-7

roots

e^{x} - 3 x - 7

x = - \frac{3 W _{- 1} ( - \frac{1}{3 e ^{\frac{7}{3}}} ) + 7}{3}, x = - \frac{3 W _{0} ( - \frac{1}{3 e ^{\frac{7}{3}}} ) + 7}{3}

Solution steps

e^{x} - 3 x - 7

The roots are the intercepts with the x-axis

(y = 0)

e^{x} - 3 x - 7 = 0

Prepare

e^{x} - 3 x - 7 = 0

for Lambert form:

1 = - e^{- x} (- 3 x - 7)

Rewrite the equation with

- x - \frac{7}{3} = u

and

x = - \frac{3 u + 7}{3}

1 = - e^{- (- \frac{3 u + 7}{3})} (- 3 (- \frac{3 u + 7}{3}) - 7)

Simplify

1 = - 3 e^{\frac{3 u + 7}{3}} u

Rewrite

1 = - 3 e^{\frac{3 u + 7}{3}} u

in Lambert form:

e^{u} u = - \frac{1}{3 e ^{\frac{7}{3}}}

Solve

e^{u} u = - \frac{1}{3 e ^{\frac{7}{3}}} : u = W_{- 1} (- \frac{1}{3 e ^{\frac{7}{3}}}), u = W_{0} (- \frac{1}{3 e ^{\frac{7}{3}}})

u = W_{- 1} (- \frac{1}{3 e ^{\frac{7}{3}}}), u = W_{0} (- \frac{1}{3 e ^{\frac{7}{3}}})

Substitute back

u = - x - \frac{7}{3},

solve for

x

Solve

- x - \frac{7}{3} = W_{- 1} (- \frac{1}{3 e ^{\frac{7}{3}}}) : x = - \frac{3 W _{- 1} ( - \frac{1}{3 e ^{\frac{7}{3}}} ) + 7}{3}

Solve

- x - \frac{7}{3} = W_{0} (- \frac{1}{3 e ^{\frac{7}{3}}}) : x = - \frac{3 W _{0} ( - \frac{1}{3 e ^{\frac{7}{3}}} ) + 7}{3}

x = - \frac{3 W _{- 1} ( - \frac{1}{3 e ^{\frac{7}{3}}} ) + 7}{3}, x = - \frac{3 W _{0} ( - \frac{1}{3 e ^{\frac{7}{3}}} ) + 7}{3}

3^{3 x + 1} = 27

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

18 e^{6 x} = 2

2^{2 x} - 3 \times 2^{x} = - 2

What is the answer to roots e^x-3x-7 ?
The answer to roots e^x-3x-7 is x=-(3\W_{-1}(-1/(3e^{7/3)})+7)/3 ,x=-(3\W_{0}(-1/(3e^{7/3)})+7)/3