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

The answer to 2+x=2^x 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=2^x

2 + x = 2^{x}

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

Solution steps

2 + x = 2^{x}

Prepare

2 + x = 2^{x}

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

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 )}

8 \cdot 2^{3 x} - 32 = 0

3^{y} = 81

5 (7^{(2 x - 1)}) = 7 (5^{(x + 2)})

3^{2 x - 1} = 93

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