What is the answer to 2^{-x}=x-9 ?

The answer to 2^{-x}=x-9 is x=(\W_{0}(\frac{ln(2))/(512))}{ln(2)}+9

Solve for x: 2^{-x}=x-9

The solution is x=(\W_{0}(\frac{ln(2))/(512))}{ln(2)}+9

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2^{- x} = x - 9

x = \frac{W _{0} ( \frac{l n ( 2 )}{512} )}{ln ( 2 )} + 9

Decimal Notation

x = 9.00195 \dots

Solution steps

2^{- x} = x - 9

Prepare

2^{- x} = x - 9

for Lambert form:

1 = (x - 9) e^{l n (2) x}

Rewrite the equation with

(x - 9) ln (2) = u

and

x = \frac{u}{ln ( 2 )} + 9

1 = ((\frac{u}{ln ( 2 )} + 9) - 9) e^{l n (2) (\frac{u}{l n ( 2 )} + 9)}

Simplify

1 = \frac{e ^{u + 9 l n (2)} u}{ln ( 2 )}

Rewrite

1 = \frac{e ^{u + 9 l n (2)} u}{ln ( 2 )}

in Lambert form:

e^{u} u = \frac{ln ( 2 )}{512}

Solve

e^{u} u = \frac{ln ( 2 )}{512} : u = W_{0} (\frac{ln ( 2 )}{512})

u = W_{0} (\frac{ln ( 2 )}{512})

Substitute back

u = (x - 9) ln (2),

solve for

x

Solve

(x - 9) ln (2) = W_{0} (\frac{ln ( 2 )}{512}) : x = \frac{W _{0} ( \frac{l n ( 2 )}{512} )}{ln ( 2 )} + 9

x = \frac{W _{0} ( \frac{l n ( 2 )}{512} )}{ln ( 2 )} + 9

2^{x - 3} = \frac{1}{8}

90 - 0.04 x = 30 e^{0.001 x}

5^{n} = 1

e^{2 l n (x)} = 225

12 5^{l o g_{3} (x + 3)} = 8

What is the answer to 2^{-x}=x-9 ?
The answer to 2^{-x}=x-9 is x=(\W_{0}(\frac{ln(2))/(512))}{ln(2)}+9
Solve for x: 2^{-x}=x-9
The solution is x=(\W_{0}(\frac{ln(2))/(512))}{ln(2)}+9