What is the answer to e^x-x=2 ?

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

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e^x-x=2

e^{x} - x = 2

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

Decimal Notation

x = 1.14619 \dots, x = - 1.84140 \dots

Solution steps

e^{x} - x = 2

Prepare

e^{x} - x = 2

for Lambert form:

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

Rewrite the equation with

- x - 2 = u

and

x = - u - 2

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

Simplify

1 = - e^{u + 2} u

Rewrite

1 = - e^{u + 2} u

in Lambert form:

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

Solve

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

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

Substitute back

u = - x - 2,

solve for

x

Solve

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

Solve

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

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

3^{x + 1} = 3^{2 x - 1} + 6

5^{4 - x} = 0

8 (4^{6 - 2 x}) + 13 = 41

7^{x} = 2 \cdot 5^{x}

6^{x} = \frac{1}{6}

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