Choose the greatest number of the lot.

(5^{- 1})^{2}

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(5^{- 2})^{- 1}

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(2^{2})^{- 5}

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(3^{- 2})^{- 1}

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Solution

The correct option is B $(5^{- 2})^{- 1}$
$a^{- m} = \frac{1}{a^{m}}$

$⟹ (5^{- 1})^{2} = (\frac{1}{5})^{2} = \frac{1}{25} = 0.04 .$

$(a^{m})^{n} = a^{m n}$

$⟹ (5^{- 2})^{- 1} = 5^{- 2 \times - 1} = 5^{2} = 25.$

Also, $(2^{2})^{- 5} = 2^{2 \times - 5} = 2^{- 10} = \frac{1}{2^{10}},$

$⟹ (3^{- 2})^{- 1} = 3^{- 2 \times - 1} = 3^{2} = 9.$

$∴ (2^{2})^{- 5} < (5^{- 1})^{2} < (3^{- 2})^{- 1} < (5^{- 2})^{- 1}$
i.e., $(5^{- 2})^{- 1}$ is the greatest of the lot.

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