What is 2^{-x}(2^{-x}+1)>= 2^x+1 ?

The solution to 2^{-x}(2^{-x}+1)>= 2^x+1 is x<= 0

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2^{-x}(2^{-x}+1)>= 2^x+1

2^{- x} (2^{- x} + 1) \geq 2^{x} + 1

x \leq 0

Interval Notation

(- \infty, 0]

Solution steps

2^{- x} (2^{- x} + 1) \geq 2^{x} + 1

Apply exponent rules

\frac{1}{2 ^{x}} + 1 \geq (2^{x})^{2} + 2^{x}

Let

u = 2^{x}

\frac{1}{u} + 1 \geq u^{2} + u

\frac{1}{u} + 1 \geq u^{2} + u : u = - 1 or 0 < u \leq 1

u = - 1 or 0 < u \leq 1

Substitute back

u = 2^{x}

2^{x} = - 1 or 0 < 2^{x} \leq 1

2^{x} = - 1 :

No Solution for

x \in R

0 < 2^{x} \leq 1 : x \leq 0

Combine the intervals

No Solution or x \leq 0

Merge Overlapping Intervals

x \leq 0

(- \frac{1}{3})^{2}

2^{6}

4 x^{2} - 36

7 (x + 8) = 49

4 x^{2} = 9

What is 2^{-x}(2^{-x}+1)>= 2^x+1 ?
The solution to 2^{-x}(2^{-x}+1)>= 2^x+1 is x<= 0