What is the answer to 2^x=x+2 ?

The answer to 2^x=x+2 is x=2,x=-(\W_{0}(-(ln(2))/4)+2ln(2))/(ln(2))

The solution is x=2,x=-(\W_{0}(-(ln(2))/4)+2ln(2))/(ln(2))

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2^x=x+2

2^{x} = x + 2

x = 2, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{4} ) + 2 ln ( 2 )}{ln ( 2 )}

Decimal Notation

x = 2, x = - 1.69009 \dots

Solution steps

2^{x} = x + 2

Prepare

2^{x} = x + 2

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

e^{u} u = - \frac{ln ( 2 )}{4}

Solve

e^{u} u = - \frac{ln ( 2 )}{4} : u = - 4 ln (2), u = W_{0} (- \frac{ln ( 2 )}{4})

u = - 4 ln (2), u = W_{0} (- \frac{ln ( 2 )}{4})

Substitute back

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

solve for

x

Solve

(- x - 2) ln (2) = - 4 ln (2) : x = 2

Solve

(- x - 2) ln (2) = W_{0} (- \frac{ln ( 2 )}{4}) : x = - \frac{W _{0} ( - \frac{l n ( 2 )}{4} ) + 2 ln ( 2 )}{ln ( 2 )}

x = 2, x = - \frac{W _{0} ( - \frac{l n ( 2 )}{4} ) + 2 ln ( 2 )}{ln ( 2 )}

15 \cdot 5^{x} = 53 \cdot 2^{x}

5^{x} = 120

x, y = e^{- 2 x}

2^{x + 7} \cdot 4^{x - 1} = 8^{2 x + 3}

3000 = 100 e^{n}

What is the answer to 2^x=x+2 ?
The answer to 2^x=x+2 is x=2,x=-(\W_{0}(-(ln(2))/4)+2ln(2))/(ln(2))
Solve for x: 2^x=x+2
The solution is x=2,x=-(\W_{0}(-(ln(2))/4)+2ln(2))/(ln(2))