What is (7^{-1}-x^{-1})^{-1^'} ?

The solution to (7^{-1}-x^{-1})^{-1^'} is (7x)/(x-7)

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simplify (7^{-1}-x^{-1})^{-1^'}

simplify

(7^{- 1} - x^{- 1})^{- 1^{^{'}}}

\frac{7 x}{x - 7}

Solution steps

(7^{- 1} - x^{- 1})^{- 1^{^{'}}}

Apply exponent rule:

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

= \frac{1}{( 7 ^{- 1} - x ^{- 1} ) ^{1^{^{^{'}}}}}

(7^{- 1} - x^{- 1})^{1^{^{'}}} = \frac{( x - 7 ) ^{1^{^{'}}}}{( 7 x ) ^{1^{^{'}}}}

= \frac{1}{\frac{( x - 7 ) ^{1^{^{^{'}}}}}{( 7 x ) ^{1^{^{^{'}}}}}}

Apply the fraction rule:

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

= \frac{( 7 x ) ^{1^{^{^{'}}}}}{( x - 7 ) ^{1^{^{^{'}}}}}

Apply rule

1^{a} = 1

1^{^{'}} = 1

= \frac{( 7 x ) ^{1}}{( x - 7 ) ^{1}}

Apply rule

a^{1} = a

= \frac{7 x}{x - 7}

(x)^{\frac{7}{12}}

b^{4} (3 a b^{6})

4 x \cdot 2 x

4.71 \cdot 1 0^{- 5}

\frac{x ^{2}}{( 2 π ) ^{2}} (4 π)

What is (7^{-1}-x^{-1})^{-1^'} ?
The solution to (7^{-1}-x^{-1})^{-1^'} is (7x)/(x-7)