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

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

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e^{3b}=2+b

e^{3 b} = 2 + b

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

Solution steps

e^{3 b} = 2 + b

Prepare

e^{3 b} = 2 + b

for Lambert form:

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

Rewrite the equation with

- 3 b - 6 = u

and

b = - \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 b - 6,

solve for

b

Solve

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

Solve

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

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

e^{4 x} = 3

1 0^{x} = 6

8^{3 x} = \frac{1}{2}

3^{x + 1} - 3^{x} = 18

2 (3^{4 x - 5}) = 7^{2 x}

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