What is the answer to e^{2x-1}=x+2 ?

The answer to e^{2x-1}=x+2 is x=-(\W_{-1}(-2/(e^5))+4)/2 ,x=-(\W_{0}(-2/(e^5))+4)/2

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

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

x = - \frac{W _{- 1} ( - \frac{2}{e ^{5}} ) + 4}{2}, x = - \frac{W _{0} ( - \frac{2}{e ^{5}} ) + 4}{2}

Decimal Notation

x = 1.05905 \dots, x = - 1.99316 \dots

Solution steps

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

Prepare

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

for Lambert form:

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

Rewrite the equation with

- 2 x - 4 = u

and

x = - \frac{u + 4}{2}

1 = ((- \frac{u + 4}{2}) + 2) e^{- 2 (- \frac{u + 4}{2}) + 1}

Simplify

1 = e^{u + 5} (- \frac{u + 4}{2} + 2)

Rewrite

1 = e^{u + 5} (- \frac{u + 4}{2} + 2)

in Lambert form:

e^{u} u = - \frac{2}{e ^{5}}

Solve

e^{u} u = - \frac{2}{e ^{5}} : u = W_{- 1} (- \frac{2}{e ^{5}}), u = W_{0} (- \frac{2}{e ^{5}})

u = W_{- 1} (- \frac{2}{e ^{5}}), u = W_{0} (- \frac{2}{e ^{5}})

Substitute back

u = - 2 x - 4,

solve for

x

Solve

- 2 x - 4 = W_{- 1} (- \frac{2}{e ^{5}}) : x = - \frac{W _{- 1} ( - \frac{2}{e ^{5}} ) + 4}{2}

Solve

- 2 x - 4 = W_{0} (- \frac{2}{e ^{5}}) : x = - \frac{W _{0} ( - \frac{2}{e ^{5}} ) + 4}{2}

x = - \frac{W _{- 1} ( - \frac{2}{e ^{5}} ) + 4}{2}, x = - \frac{W _{0} ( - \frac{2}{e ^{5}} ) + 4}{2}

e^{x} = 5275

50 = 40 - 80 e^{(- 0.2 t)}

2^{(- x + 3)} = 64

e^{t} = - 1

9^{x - 1} - 2 \cdot 3^{x} - 27 = 0

What is the answer to e^{2x-1}=x+2 ?
The answer to e^{2x-1}=x+2 is x=-(\W_{-1}(-2/(e^5))+4)/2 ,x=-(\W_{0}(-2/(e^5))+4)/2