What is the answer to e^{-x}+1/5 x-1=0 ?

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

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e^{-x}+1/5 x-1=0

e^{- x} + \frac{1}{5} x - 1 = 0

x = 0, x = W_{0} (- \frac{5}{e ^{5}}) + 5

Decimal Notation

x = 0, x = 4.96511 \dots

Solution steps

e^{- x} + \frac{1}{5} x - 1 = 0

Prepare

e^{- x} + \frac{1}{5} x - 1 = 0

for Lambert form:

1 = - e^{x} (\frac{1}{5} x - 1)

Rewrite the equation with

x - 5 = u

and

x = u + 5

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

Simplify

1 = - \frac{1}{5} e^{u + 5} u

Rewrite

1 = - \frac{1}{5} e^{u + 5} u

in Lambert form:

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

Solve

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

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

Substitute back

u = x - 5,

solve for

x

Solve

x - 5 = - 5 : x = 0

Solve

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

x = 0, x = W_{0} (- \frac{5}{e ^{5}}) + 5

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3 6^{2 x} = \frac{1}{6 ^{2 (x + 1)}}

7^{x^{2}} = 7^{6 - x}

2^{x + 6} - 2^{x + 2} = 1

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