What is {[(2/3)^{-2}]^{-1}}^{-1} ?

The solution to {[(2/3)^{-2}]^{-1}}^{-1} is 9/4

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simplify {[(2/3)^{-2}]^{-1}}^{-1}

simplify

⎩ ⎨ ⎧ [(\frac{2}{3})^{- 2}]^{- 1} ⎭ ⎬ ⎫^{- 1}

\frac{9}{4}

Decimal Notation

2.25

Solution steps

⎩ ⎨ ⎧ [(\frac{2}{3})^{- 2}]^{- 1} ⎭ ⎬ ⎫^{- 1}

[(\frac{2}{3})^{- 2}]^{- 1} : (\frac{2}{3})^{2}

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

Apply exponent rule:

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

(\frac{2}{3})^{2} = \frac{2 ^{2}}{3 ^{2}}

= \frac{1}{\frac{2 ^{2}}{3 ^{2}}}

Apply the fraction rule:

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

= \frac{3 ^{2}}{2 ^{2}}

3^{2} = 9

= \frac{9}{2 ^{2}}

2^{2} = 4

= \frac{9}{4}

(a s + b s) (a + b)^{- 1}

e^{- 10000000000}

5.401 \cdot 1 0^{- 1}

9.5 \cdot 1 0^{- 4}

What is {[(2/3)^{-2}]^{-1}}^{-1} ?
The solution to {[(2/3)^{-2}]^{-1}}^{-1} is 9/4