What is the answer to solvefor y,3x^3y+e^y=4x+6 ?

The answer to solvefor y,3x^3y+e^y=4x+6 is y=-(3\W(1/(3e^{\frac{-4x-6){3x^3}}x^3})x^3-4x-6)/(3x^3)

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solvefor y,3x^3y+e^y=4x+6

solve for

y, 3 x^{3} y + e^{y} = 4 x + 6

y = - \frac{3 W ( \frac{1}{3 e ^{\frac{- 4 x - 6}{3 x ^{3}}} x ^{3}} ) x ^{3} - 4 x - 6}{3 x ^{3}}

Solution steps

3 x^{3} y + e^{y} = 4 x + 6

Prepare

3 x^{3} y + e^{y} = 4 x + 6

for Lambert form:

1 = (4 x + 6 - 3 x^{3} y) e^{- y}

Rewrite the equation with

- y + \frac{4 x + 6}{3 x ^{3}} = u

and

y = - \frac{3 x ^{3} u - 4 x - 6}{3 x ^{3}}

1 = (4 x + 6 - 3 x^{3} (- \frac{3 x ^{3} u - 4 x - 6}{3 x ^{3}})) e^{- (- \frac{3 x ^{3} u - 4 x - 6}{3 x ^{3}})}

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

e^{u} u = \frac{1}{3 e ^{\frac{- 4 x - 6}{3 x ^{3}}} x ^{3}} : u = W (\frac{1}{3 e ^{\frac{- 4 x - 6}{3 x ^{3}}} x ^{3}})

u = W (\frac{1}{3 e ^{\frac{- 4 x - 6}{3 x ^{3}}} x ^{3}})

Substitute back

u = - y + \frac{4 x + 6}{3 x ^{3}},

solve for

y

Solve

- y + \frac{4 x + 6}{3 x ^{3}} = W (\frac{1}{3 e ^{\frac{- 4 x - 6}{3 x ^{3}}} x ^{3}}) : y = - \frac{3 W ( \frac{1}{3 e ^{\frac{- 4 x - 6}{3 x ^{3}}} x ^{3}} ) x ^{3} - 4 x - 6}{3 x ^{3}}

y = - \frac{3 W ( \frac{1}{3 e ^{\frac{- 4 x - 6}{3 x ^{3}}} x ^{3}} ) x ^{3} - 4 x - 6}{3 x ^{3}}

2^{x^{2} + 3 x - 10} = 256

4^{3 x} = 2^{2 x - 8}

3^{x - 1} = 81

6 4^{x} = 0.5

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

What is the answer to solvefor y,3x^3y+e^y=4x+6 ?
The answer to solvefor y,3x^3y+e^y=4x+6 is y=-(3\W(1/(3e^{\frac{-4x-6){3x^3}}x^3})x^3-4x-6)/(3x^3)