How do you write $18 f^{- 1}$ $18f^-1$ using only positive exponents?

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Answer 1 · 2017-03-20T07:14:29

$18 f^{- 1} = \frac{18}{f}$ $18f^(-1)=18/f$

As $a^{- m} = \frac{1}{a^{m}}$ $a^(-m)=1/a^m$

$f^{- 1} = \frac{1}{f^{1}} = \frac{1}{f}$ $f^(-1)=1/(f^1)=1/f$

Hence $18 f^{- 1} = 18 \times \frac{1}{f} = \frac{18}{f}$ $18f^(-1)=18xx1/f=18/f$

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