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Popular >

2^x<= x+1

Solution

2^{x} \leq x + 1

Solution

x \leq 0 or x \geq 1

+1

Solution steps

2^{x} \leq x + 1

Prepare

2^{x} \leq x + 1

for Lambert form:

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

Rewrite the equation with

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

and

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

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

Simplify

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

Rewrite

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

in Lambert form:

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

Solve

e^{u} u \geq - \frac{ln ( 2 )}{2} : u \leq - 2 ln (2) or u \geq - ln (2)

u \leq - 2 ln (2) or u \geq - ln (2)

Substitute back

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

solve for

x

Solve

(- x - 1) ln (2) \leq - 2 ln (2) or (- x - 1) ln (2) \geq - ln (2) : x \leq 0 or x \geq 1

x \leq 0 or x \geq 1

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Frequently Asked Questions (FAQ)

What is 2^x<= x+1 ?
The solution to 2^x<= x+1 is x<= 0\lor x>= 1