What is the answer to 2^x-x-1=0 ?

The answer to 2^x-x-1=0 is x=1,x=0

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2^x-x-1=0

2^{x} - x - 1 = 0

x = 1, x = 0

Solution steps

2^{x} - x - 1 = 0

Prepare

2^{x} - x - 1 = 0

for Lambert form:

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

Rewrite the equation with

(- x - 1) ln (2) = u

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

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

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

Substitute back

u = (- x - 1) ln (2),

solve for

x

Solve

(- x - 1) ln (2) = - 2 ln (2) : x = 1

Solve

(- x - 1) ln (2) = - ln (2) : x = 0

x = 1, x = 0

x, 3456 = 2^{x} 3^{y}

2^{x} = 22

3^{a - 7} = 2 7^{2 a}

3^{2 x} = \frac{1}{81}

e^{2 - 3 x} = 12