Hello, Tresa332006!
When you move the negatives what steps did you take
or did you just move top to bottom or bottom to top?
Yes, that's what I did . . . but there's a reason behind it.
By defintion: \(\displaystyle \L\,a^{-2}\:=\:\frac{1}{a^2}\)
So a factor in the numerator with a negative exponent
\(\displaystyle \;\;\)can be moved "downstairs". \(\displaystyle \;\)(Remember to change the sign.)
What about a negative exponent in the denominator?\(\displaystyle \L\;\frac{1}{a^{-3}}\)
This means: \(\displaystyle \L\,\frac{1}{\left(\frac{1}{a^3}\right)}\;\) . . . which is a division problem: \(\displaystyle \L\,1\,\div\frac{1}{a^3}\)
"Invert and multiply": \(\displaystyle \L\:1\,\times\,\frac{a^3}{1}\;=\;a^3\)
Therefore: \(\displaystyle \L\,\frac{1}{a^{-3}}\:=\:a^3\)
\(\displaystyle \;\;\)(A negative exponent in the denominator can be moved "upstairs".)
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Here's another situation that drives people crazy: \(\displaystyle \L\,\left(\frac{a}{b}\right)^{-2}\)
After simplifying this once, you should be able to come up with your own Rule.
We have: \(\displaystyle \L\,\frac{a^{-2}}{b^{-2}}\;=\;\frac{b^2}{a^2} \;= \;\left(\frac{b}{a}\right)^2\)
Do you see it? . . . \(\displaystyle \L\,\left(\frac{a}{b}\right)^{-2}\;=\;\left(\frac{b}{a}\right)^2\)
That's right . . . A negative exponent on a fraction
flips the fraction.
