What is the answer to 2e^x(x-1)+1=0 ?

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

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2e^x(x-1)+1=0

2 e^{x} (x - 1) + 1 = 0

x = W_{- 1} (- \frac{1}{2 e}) + 1, x = W_{0} (- \frac{1}{2 e}) + 1

Decimal

x = - 1.67834 \dots, x = 0.76803 \dots

Solution steps

2 e^{x} (x - 1) + 1 = 0

Prepare

2 e^{x} (x - 1) + 1 = 0

for Lambert form:

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

Rewrite the equation with

x - 1 = u

and

x = u + 1

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = - \frac{1}{2 e}

Solve

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

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

Substitute back

u = x - 1,

solve for

x

Solve

x - 1 = W_{- 1} (- \frac{1}{2 e}) : x = W_{- 1} (- \frac{1}{2 e}) + 1

Solve

x - 1 = W_{0} (- \frac{1}{2 e}) : x = W_{0} (- \frac{1}{2 e}) + 1

x = W_{- 1} (- \frac{1}{2 e}) + 1, x = W_{0} (- \frac{1}{2 e}) + 1

5^{5} = 312 5^{3 x + 1}

2^{2 x} - 5 \times 2^{x} + 4 = 0

4^{3 x} + 8^{2 x + 1} = 1 6^{x}

e^{6 x} = 0

1 0^{7 - x} = 5^{x + 1}

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