# Simplify an Expression with Negative Exponents

• Jun 1st 2008, 10:58 AM
needymathperson
Simplify an Expression with Negative Exponents

(ab)^-1 / (a^-2 + b^-2)

if you could explain it with an appropriate property, that would be great!!

thanks..!!
• Jun 1st 2008, 12:21 PM
masters
$\displaystyle \frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2 }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2} {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}$

I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.
• Jun 1st 2008, 12:26 PM
topsquark
Quote:

Originally Posted by masters
$\displaystyle \frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{\frac{1}{ab}}{\frac{1}{a^2}+\frac{1}{b^2 }}=\frac{\frac{1}{ab}}{\frac{b^2+a^2}{a^2b^2}}=\fr ac{1}{ab}\cdot\frac{a^2b^2}{a^2+b^2}=\frac{a^2b^2} {ab(a^2+b^2)}=\frac{ab}{a^2+b^2}$

I think that's right. Latex notation was brutal on this one...for me, anyway. I welcome others to detect any errors.

Looks good to me, but don't forget to specify that $\displaystyle a \neq 0$ and $\displaystyle b \neq 0$ independently.

-Dan
• Jun 1st 2008, 12:30 PM
Moo
Hi !

A slightly different approach :)

$\displaystyle \frac{(ab)^{-1}}{a^{-2}+b^{-2}}=\frac{1}{(ab)(a^{-2}+b^{-2})}$

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Using the power rule : $\displaystyle a^b a^c=a^{b+c}$

--> $\displaystyle (ab)a^{-2}=ba^{-1}$
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$\displaystyle =\frac{1}{ba^{-1}+ab^{-1}}=\frac{1}{\frac ba+\frac ab}=\frac{1}{\frac{b^2+a^2}{ab}}=\frac{ab}{a^2+b^2 }$

(Crying) one extra equality
• Jun 1st 2008, 12:36 PM
masters
Always keep it interesting, Moo. Good job!!
• Jun 1st 2008, 12:44 PM
needymathperson
thank you both for your help! (Clapping)
• Jun 2nd 2008, 04:34 AM
Soroban
Hello, needymathperson!

One different step . . .

Quote:

$\displaystyle \frac{(ab)^{-1}}{a^{-2} + b^{-2}}$
We have: .$\displaystyle \frac{(ab)^{-1}} {a^{-2} + b^{-2}} \;\;=\;\;\frac{\dfrac{1}{ab}} {\dfrac{1}{a^2} + \dfrac{1}{b^2}}$
Multiply top and bottom by $\displaystyle a^2b^2\!:\;\;\frac{a^2b^2\left(\dfrac{1}{ab}\right )}{a^2b^2\left(\dfrac{1}{a^2} + \dfrac{1}{b^2}\right)} \;\;=\;\;\boxed{\frac{ab}{b^2+a^2}}$