What is the answer to e^{3x}=2+x ?

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

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e^{3x}=2+x

e^{3 x} = 2 + x

x = - \frac{W _{- 1} ( - \frac{3}{e ^{6}} ) + 6}{3}, x = - \frac{W _{0} ( - \frac{3}{e ^{6}} ) + 6}{3}

Solution steps

e^{3 x} = 2 + x

Prepare

e^{3 x} = 2 + x

for Lambert form:

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

Rewrite the equation with

- 3 x - 6 = u

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = - \frac{3}{e ^{6}}

Solve

e^{u} u = - \frac{3}{e ^{6}} : u = W_{- 1} (- \frac{3}{e ^{6}}), u = W_{0} (- \frac{3}{e ^{6}})

u = W_{- 1} (- \frac{3}{e ^{6}}), u = W_{0} (- \frac{3}{e ^{6}})

Substitute back

u = - 3 x - 6,

solve for

x

Solve

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

Solve

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

x = - \frac{W _{- 1} ( - \frac{3}{e ^{6}} ) + 6}{3}, x = - \frac{W _{0} ( - \frac{3}{e ^{6}} ) + 6}{3}

e^{2 t + 1} = 9 e^{1 - t}

x - 3^{- x} = 0

4^{x} + 6^{x} = 2 \cdot 9^{x}

2^{x^{2} + x} = 4^{x + 1}

1 6^{2 x + 1} = 3^{3 x - 4}

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