What is (a^{-3}b^{-1})^{-2^'} ?

The solution to (a^{-3}b^{-1})^{-2^'} is (a^3b)^{2^'}

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simplify (a^{-3}b^{-1})^{-2^'}

simplify

(a^{- 3} b^{- 1})^{- 2^{^{'}}}

(a^{3} b)^{2^{'}}

Solution steps

(a^{- 3} b^{- 1})^{- 2^{^{'}}}

Apply exponent rule:

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

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

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

= \frac{1}{\frac{1 ^{2^{^{^{'}}}}}{( a ^{3} b ) ^{2^{^{^{'}}}}}}

Apply the fraction rule:

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

= \frac{( a ^{3} b ) ^{2^{^{^{'}}}}}{1 ^{2^{^{^{'}}}}}

Apply rule

1^{a} = 1

1^{2^{^{'}}} = 1

= \frac{( a ^{3} b ) ^{2^{^{^{'}}}}}{1}

Apply the fraction rule:

\frac{a}{1} = a

= (a^{3} b)^{2^{^{^{'}}}}

5.84 x 1 0^{- 7} N

3.13 \cdot 1 0^{- 5}

x^{2} \cdot e^{- y^{2}}

co m^{70} He ll o w

3.13 \cdot 1 0^{- 4}

What is (a^{-3}b^{-1})^{-2^'} ?
The solution to (a^{-3}b^{-1})^{-2^'} is (a^3b)^{2^'}