What is the answer to x9=8+x+e^x ?

The answer to x9=8+x+e^x is x=-\W_{-1}(-e/8)+1,x=-\W_{0}(-e/8)+1

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x9=8+x+e^x

x 9 = 8 + x + e^{x}

x = - W_{- 1} (- \frac{e}{8}) + 1, x = - W_{0} (- \frac{e}{8}) + 1

Decimal Notation

x = 2.45323 \dots, x = 1.65250 \dots

Solution steps

x \cdot 9 = 8 + x + e^{x}

Prepare

x 9 = 8 + x + e^{x}

for Lambert form:

1 = (8 x - 8) e^{- x}

Rewrite the equation with

- x + 1 = u

and

x = - u + 1

1 = (8 (- u + 1) - 8) e^{- (- u + 1)}

Simplify

1 = - 8 e^{u - 1} u

Rewrite

1 = - 8 e^{u - 1} u

in Lambert form:

e^{u} u = - \frac{e}{8}

Solve

e^{u} u = - \frac{e}{8} : u = W_{- 1} (- \frac{e}{8}), u = W_{0} (- \frac{e}{8})

u = W_{- 1} (- \frac{e}{8}), u = W_{0} (- \frac{e}{8})

Substitute back

u = - x + 1,

solve for

x

Solve

- x + 1 = W_{- 1} (- \frac{e}{8}) : x = - W_{- 1} (- \frac{e}{8}) + 1

Solve

- x + 1 = W_{0} (- \frac{e}{8}) : x = - W_{0} (- \frac{e}{8}) + 1

x = - W_{- 1} (- \frac{e}{8}) + 1, x = - W_{0} (- \frac{e}{8}) + 1

8^{- 9^{x^{- 1}}} = \frac{1}{2}

1.005 7^{4 t} = 2.22717

0. 3^{x} = 4^{x^{2}}

4^{2 x - 6} = 48

432 = 2^{4} \cdot 3^{x}

What is the answer to x9=8+x+e^x ?
The answer to x9=8+x+e^x is x=-\W_{-1}(-e/8)+1,x=-\W_{0}(-e/8)+1