What is [{(2^{-1})^{-1}}^{-1}]^{-1} ?

The solution to [{(2^{-1})^{-1}}^{-1}]^{-1} is 2

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simplify [{(2^{-1})^{-1}}^{-1}]^{-1}

simplify

[{(2^{- 1})^{- 1}}^{- 1}]^{- 1}

2

Solution steps

[{(2^{- 1})^{- 1}}^{- 1}]^{- 1}

(2^{- 1})^{- 1} : 2

= [{2}^{- 1}]^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

= \frac{1}{\frac{1}{2}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{2}{1}

Apply the fraction rule:

\frac{a}{1} = a

= 2

10.5 \cdot 1 0^{- 3}

x d x

- 2^{- t} ln (2) (cos (2 π t))

2.7 \cdot 1 0^{- 6}

- \frac{1}{4} e^{- 4 (0)}

What is [{(2^{-1})^{-1}}^{-1}]^{-1} ?
The solution to [{(2^{-1})^{-1}}^{-1}]^{-1} is 2