Simplifying Fractions with Negative Exponents.

[alchmstlva]

I am having trouble simplifying fraction equations that include negative exponents. Here's one of the problems I'm having difficulty with:

(12y^-3/3y^2)^-2 The book says the answer works out to y^10/16 but I would like to learn how to do these problems on my own. Thanks so much for the help!

 

[tkhunny]

Use your fundamental principles and their implications.

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

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

\(\displaystyle \left(\frac{x}{y}\right)^{-1} = \left(\frac{y}{x}\right)\)

And various other useful expressions.

 

[alchmstlva]

Thanks! That really helped! We're currently doing a review but we never got a chance to go back over these in class.

 

[lookagain]

:

(12y^-3/3y^2)^-2 The book says the answer works out to y^10/16 but I ...

alchmstlva,

the expression, based on your answer, is not typed correctly.
You at least need grouping symbols around "3y^2" because of
the Order of Operations.


{[12y^(-3)]/[3y^2]}^(-2) would be one of the appropriate ways.

 

[Jason76]

\(\displaystyle (\dfrac{12y^{-3}}{3y^{2}})^{-2}\)

So this would involve dividing 12 by 3 and subtracting 2 from -3 and then negatively squaring the whole thing by -2. Is that right? Sorry about typesetting mistake, I'm trying to learn La Tex.

 

[mmm4444bot]

\(\displaystyle \dfrac{12y^{-3}}{3y^2}\)

Jason, that's not correct. What are you doing here, anyway? You have work to complete in your thread. :-?

 

[Jason76]

Jason, that's not correct. What are you doing here, anyway? You have work to complete in your thread. :-?

This caught my eye and I need to know this also.

 

[mmm4444bot]

\(\displaystyle (\dfrac{12y^{-3}}{3y^{2}})^{-2}\)

So this would involve dividing 12 by 3 and subtracting [the exponents] 2 from -3 [to get the simplified power of y inside the parentheses] and then negatively squaring the whole thing by -2. Is that right?

Sorry about typesetting mistake, I'm trying to learn La Tex.

Oh, I get what happened now. Your partial LaTex was a work in progress, and you were probably adding the text above while I was viewing the first attempt.

Yes, you are mostly correct. Please explain your meaning of "negatively squaring by -2" ? :-?


LaTex Hint: Note my use of \left( and \right) below.

Right-click and follow sub-menus to see the Tex Commands.

\(\displaystyle \left( \dfrac{12y^{-3}}{3y^{2}} \right)^{-2}\)

 

[Jason76]

Oh, I get what happened now. Your partial LaTex was a work in progress, and you were probably adding the text above while I was viewing the first attempt.

Yes, you are mostly correct. Please explain your meaning of "negatively squaring by -2" ? :-?


LaTex Hint: Note my use of \left( and \right) below.

Right-click and follow sub-menus to see the Tex Commands.

\(\displaystyle \left( \dfrac{12y^{-3}}{3y^{2}} \right)^{-2}\)

From what I can see the answer should be

\(\displaystyle 4y^{10}\)

because the exponents subtract to -5, and the main numbers divide to 4. Next you multiply the exponents of -5 and -2 (which lies outside the parenthesis).

 

[lookagain]

From what I can see the answer should be \(\displaystyle 4y^{10}\)because the exponents subtract to -5,
and the main numbers divide to 4.
Next you multiply the exponents of -5 and -2 (which lies outside the parenthesis).

No, every factor in the product inside the grouping symbols is raised to outermost exponent.


\(\displaystyle \bigg(\dfrac{12y^{-3}}{3y^2}\bigg)^{-2} \ = \)


\(\displaystyle \bigg(\dfrac{4y^{-3}}{y^2}\bigg)^{-2} \ = \)


\(\displaystyle \bigg(4^{-2}\bigg)\bigg(\dfrac{y^{-3}}{y^2}\bigg)^{-2}\)



Then continue...