What is the answer to x^2-2xe^{-x}+e^{-2x}=0 ?

The answer to x^2-2xe^{-x}+e^{-2x}=0 is x=\W_{0}(1)

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x^2-2xe^{-x}+e^{-2x}=0

x^{2} - 2 x e^{- x} + e^{- 2 x} = 0

x = W_{0} (1)

Decimal Notation

x = 0.56714 \dots

Solution steps

x^{2} - 2 x e^{- x} + e^{- 2 x} = 0

Factor

x^{2} - 2 x e^{- x} + e^{- 2 x} : (x - e^{- x})^{2}

(x - e^{- x})^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - e^{- x} = 0

Solve

x - e^{- x} = 0 : x = W_{0} (1)

x = W_{0} (1)

3 2^{2 x - 3} = \frac{1}{8}

2^{x + 3} = 5^{x}

3^{x} + \frac{4}{3 ^{x}} = 5

9^{x} + 3^{x + 1} = 28

What is the answer to x^2-2xe^{-x}+e^{-2x}=0 ?
The answer to x^2-2xe^{-x}+e^{-2x}=0 is x=\W_{0}(1)