What is the answer to 2^{-2x}+2^{-x}=2 ?

The answer to 2^{-2x}+2^{-x}=2 is x=0

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2^{-2x}+2^{-x}=2

2^{- 2 x} + 2^{- x} = 2

x = 0

Solution steps

2^{- 2 x} + 2^{- x} = 2

Apply exponent rules

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

Rewrite the equation with

2^{x} = u

(u)^{- 2} + (u)^{- 1} = 2

Solve

u^{- 2} + u^{- 1} = 2 : u = 1, u = - \frac{1}{2}

u = 1, u = - \frac{1}{2}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 1 : x = 0

Solve

2^{x} = - \frac{1}{2} :

No Solution for

x \in R

x = 0

5^{2 - x} = \frac{1}{25}

3^{2 x + 1} \cdot 3^{x} = 243

6^{5} = 6^{x} \cdot 6^{4}

8 1^{2 p + 3} \cdot 27 = \frac{1}{27}

2^{2 x + 1} - 5 (2^{x}) + 2 = 0