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

The solution to pi^{1+x}2^{2-2x}(2pi)^{-1} is 2^{-2x+1}pi^x

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simplify pi^{1+x}2^{2-2x}(2pi)^{-1}

simplify

π^{1 + x} 2^{2 - 2 x} (2 π)^{- 1}

2^{- 2 x + 1} π^{x}

Solution steps

π^{1 + x} \cdot 2^{2 - 2 x} (2 π)^{- 1}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(2 π)^{- 1} = 2^{- 1} π^{- 1}

= 2^{- 1} π^{- 1} \cdot 2^{- 2 x + 2} π^{x + 1}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

π^{1 + x} π^{- 1} = π^{1 + x - 1}

= 2^{2 - 2 x} \cdot 2^{- 1} π^{1 + x - 1}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

2^{2 - 2 x} \cdot 2^{- 1} = 2^{2 - 2 x - 1}

= 2^{2 - 2 x - 1} π^{1 + x - 1}

Simplify

2 - 2 x - 1 : - 2 x + 1

= 2^{- 2 x + 1} π^{x + 1 - 1}

Simplify

1 + x - 1 : x

= 2^{- 2 x + 1} π^{x}

e^{- x t}

5.5 \cdot 1 0^{- 5}

4 e^{- x} cos (x)

2 (3)^{2}

(3 x^{3}) (2 x^{2})

What is pi^{1+x}2^{2-2x}(2pi)^{-1} ?
The solution to pi^{1+x}2^{2-2x}(2pi)^{-1} is 2^{-2x+1}pi^x