What is the answer to e^{-x}(1+x)=0.005 ?

The answer to e^{-x}(1+x)=0.005 is x=-\W_{-1}(-(0.005)/e)-1,x=-\W_{0}(-(0.005)/e)-1

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e^{-x}(1+x)=0.005

e^{- x} (1 + x) = 0.005

x = - W_{- 1} (- \frac{0.005}{e}) - 1, x = - W_{0} (- \frac{0.005}{e}) - 1

Decimal Notation

x = 7.43012 \dots, x = - 0.99815 \dots

Solution steps

e^{- x} (1 + x) = 0.005

Rewrite the equation with

- x - 1 = u

and

x = - u - 1

e^{- (- u - 1)} (1 + (- u - 1)) = 0.005

Simplify

- e^{u + 1} u = 0.005

Rewrite

- e^{u + 1} u = 0.005

in Lambert form:

e^{u} u = - \frac{0.005}{e}

Solve

e^{u} u = - \frac{0.005}{e} : u = W_{- 1} (- \frac{0.005}{e}), u = W_{0} (- \frac{0.005}{e})

u = W_{- 1} (- \frac{0.005}{e}), u = W_{0} (- \frac{0.005}{e})

Substitute back

u = - x - 1,

solve for

x

Solve

- x - 1 = W_{- 1} (- \frac{0.005}{e}) : x = - W_{- 1} (- \frac{0.005}{e}) - 1

Solve

- x - 1 = W_{0} (- \frac{0.005}{e}) : x = - W_{0} (- \frac{0.005}{e}) - 1

x = - W_{- 1} (- \frac{0.005}{e}) - 1, x = - W_{0} (- \frac{0.005}{e}) - 1

e^{3 x} = 2 + b

e^{2 t} - 3 e^{t} = 0

5 = 2 5^{x}

32 = 2^{p}

5 e^{4 x} = 8

What is the answer to e^{-x}(1+x)=0.005 ?
The answer to e^{-x}(1+x)=0.005 is x=-\W_{-1}(-(0.005)/e)-1,x=-\W_{0}(-(0.005)/e)-1