Multiplying/dividing with exponents

**allyourbass2212** · June 9th 2009, 05:34 PM

I am having difficulties figuring out what exactly the author of my Algebra book is doing in regards to step 2.

Expression
$(\frac{x^3}{y^2}^5) (\frac{x}{y})^{-2}$

Step 1= $\frac{x^{15}}{y^{10}} * \frac{y^2}{x^2}$

Step 2= $x^{15-2} y^{2-10}$

Could some one please clearly explain whats going on from step 1 to step 2. When the expression turns into $x^{15-2} y^{2-10}$ is where im the most confused, if someone would please explain this transition that would be greatly appreciated. Unless the author is using $\frac{a^m}{a^n}=a^{n*m}$ to combine BOTH fractions.

Thanks!

**Shyam** · June 9th 2009, 05:44 PM

Originally Posted by allyourbass2212

I am having difficulties figuring out what exactly the author of my Algebra book is doing in regards to step 2.

Expression
$(\frac{x^3}{y^2}^5) (\frac{x}{y})^{-2}$

Step 1= $\frac{x^{15}}{y^{10}} * \frac{y^2}{x^2}$

Step 2= $x^{15-2} y^{2-10}$

Could some one please clearly explain whats going on from step 1 to step 2. When the expression turns into $x^{15-2} y^{2-10}$ is where im the most confused, if someone would please explain this transition that would be greatly appreciated. Unless the author is using $\frac{a^m}{a^n}=a^{n*m}$ to combine BOTH fractions.

Thanks!

Look for exponent laws:
Exponent Laws - Learning Algebra Fundamentals

press the green button every time.

**allyourbass2212** · June 9th 2009, 05:47 PM

Yes I understand when its a single fraction using $\frac{a^m}{a^n}=a^{m-n}$ . I am familiar with the exponent laws and in fact I have them laying out in front of me. I am just uncertain whats going on in this particular situation.

**cmf0106** · June 9th 2009, 05:54 PM

I am no math expert, but I think this might help clarify.

Instead of seeing it like this
$\frac{x^{15}}{y^{10}} * \frac{y^2}{x^2}$

View it as this
$\frac{x^{15}y^2}{y^{10}x^2}$

Then just use the exponent property $\frac{a^m}{a^n}=a^{m-n}$ to solve the rest.

e.g.

15-2, 2-10 resulting in $x^{13}y^{-8}$
or $\frac{x^{13}}{y^8}$ if you want to remove negative exponents using the property $a^{-n}=\frac{1}{a^n}$

But again im no expert at math, I would have someone else verify this statement. I would hate to be the source of any confusion.

**Shyam** · June 9th 2009, 05:59 PM

Originally Posted by allyourbass2212

Yes I understand when its a single fraction using $\frac{a^m}{a^n}=a^{m-n}$ . I am familiar with the exponent laws and in fact I have them laying out in front of me. I am just uncertain whats going on in this particular situation.

$ \frac{x^{15}}{y^{10}} \;\times\;\frac{y^2}{x^2}$

$ \frac{x^{15}}{x^2}\;\times\;\frac{y^2}{y^{10}} $

$ x^{15-2}\;\times \;y^{2-10} $

$ x^{13}\;\times \;y^{-8} $

$ x^{13}\;\times \;\frac{1}{y^{8}} $

$ \frac{x^{13}}{y^{8}} $

**allyourbass2212** · June 9th 2009, 06:12 PM

Thanks both of you.

I think the way cmf0106 set it up is easier to understand though... can anyone confirm he did it correctly?

Thanks again guys.

**Shyam** · June 9th 2009, 06:33 PM

Originally Posted by allyourbass2212

Thanks both of you.

I think the way cmf0106 set it up is easier to understand though... can anyone confirm he did it correctly?

Thanks again guys.

yes, he did correctly.

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