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

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

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e^{2x}=3+x

e^{2 x} = 3 + x

x = - \frac{W _{- 1} ( - \frac{2}{e ^{6}} ) + 6}{2}, x = - \frac{W _{0} ( - \frac{2}{e ^{6}} ) + 6}{2}

Decimal Notation

x = 0.64694 \dots, x = - 2.99750 \dots

Solution steps

e^{2 x} = 3 + x

Prepare

e^{2 x} = 3 + x

for Lambert form:

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

Rewrite the equation with

- 2 x - 6 = u

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = - \frac{2}{e ^{6}}

Solve

e^{u} u = - \frac{2}{e ^{6}} : u = W_{- 1} (- \frac{2}{e ^{6}}), u = W_{0} (- \frac{2}{e ^{6}})

u = W_{- 1} (- \frac{2}{e ^{6}}), u = W_{0} (- \frac{2}{e ^{6}})

Substitute back

u = - 2 x - 6,

solve for

x

Solve

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

Solve

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

x = - \frac{W _{- 1} ( - \frac{2}{e ^{6}} ) + 6}{2}, x = - \frac{W _{0} ( - \frac{2}{e ^{6}} ) + 6}{2}

x^{l o g_{2} (x)} = 2

4^{x - 5} = (\frac{1}{8})^{8 - x}

\frac{1}{16} = 6 4^{4 x - 3}

e^{x} (2 x + x^{2}) = 0

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

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