What is the answer to sqrt(3x+3)=e^x ?

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

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sqrt(3x+3)=e^x

3 x + 3 = e^{x}

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

Decimal Notation

x = 0.85944 \dots, x = - 0.95015 \dots

Solution steps

3 x + 3 = e^{x}

Prepare

3 x + 3 = e^{x}

for Lambert form:

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

Rewrite the equation with

- 2 x - 2 = u

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

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

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

Substitute back

u = - 2 x - 2,

solve for

x

Solve

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

Solve

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

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

Verify Solutions:

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

True

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

True

The solutions are

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

Verify Solutions:

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

True

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

True

The solutions are

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

e^{1 - 2 x} = 4

(\frac{1}{6})^{x} = 216

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

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

(8^{n})^{3} = 8^{6}

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