# "Simplify [ (w^3) / (w^7) ]^3" (I get 3/7.) : (

#### helpmepls!

##### New member
1. Simplify and express all answers in terms of positive exponents.

. . . .$$\displaystyle \mbox{a) }\, \left(\dfrac{w^3}{w^7}\right)^3 = \dfrac{w^9}{w^{21}} = \dfrac{1}{w^{12}}$$

I want to know how you get the answer? because when I simplify I get 3/7. thank you!!

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#### ksdhart2

##### Senior Member
Well, I can't be sure because you didn't actually show any of your work, but my best guess is that saw there was a w in both the numerator and denominator and cancelled them. However, that was not a mathematically valid step. The two rules of exponents needed to solve this problem are:

$$\displaystyle \dfrac{w^n}{w^m}=w^{n-m}$$ and $$\displaystyle w^{-n}=\dfrac{1}{w^n}$$

Meaning that in the context of this specific problem:

$$\displaystyle \dfrac{w^9}{w^{21}}=w^{9-21}=w^{-12}=\dfrac{1}{w^{12}}$$

Now, personally I find rules and formulas, in and of themselves, to be fairly unhelpful, because such a rule or formula often very quickly becomes some magical, mystical thing you memorize and it works but you'll be darned if you know how or why. Instead, I find that a much better way to approach it is to think about what the concept really means, and by doing so you'll actually derive the formula or rule.

This problem deals with powers, so let's review what a power really is and what it means. Raising a number to some power is essentially just repeated multiplication. So, $$\displaystyle w^n$$ is really $$\displaystyle w \cdot w \cdot w \cdot ... \cdot w$$ with n copies of w. Accordingly, we can rewrite the expression in the penultimate step:

$$\displaystyle \dfrac{w^9}{w^{21}}=\dfrac{w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w}{w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot w}$$

Try continuing from here. What do you get after cancel out all of the common terms? How does that relate to the given rules?

#### MarkFL

##### Super Moderator
Staff member
...Now, personally I find rules and formulas, in and of themselves, to be fairly unhelpful, because such a rule or formula often very quickly becomes some magical, mystical thing you memorize and it works but you'll be darned if you know how or why. Instead, I find that a much better way to approach it is to think about what the concept really means, and by doing so you'll actually derive the formula or rule...
Excellent philosophy, one I wish more secondary school educators would promote.

#### Harry_the_cat

##### Senior Member
Excellent philosophy, one I wish more secondary school educators would promote.
Totally agree! My mantra with exponents "If in doubt, expand it out!"

#### MarkFL

##### Super Moderator
Staff member
I want to know how you get the answer?...
On a side note, I do want to say that it would be better to use descriptive thread titles. A good thread title briefly describes the question being asked, so that those providing help can see at a glance what the problem is about. It's also better for site SEO.

#### stapel

##### Super Moderator
Staff member
1. Simplify and express all answers in terms of positive exponents.

. . . .$$\displaystyle \mbox{a) }\, \left(\dfrac{w^3}{w^7}\right)^3 = \dfrac{w^9}{w^{21}} = \dfrac{1}{w^{12}}$$

I want to know how you get the answer? because when I simplify I get 3/7. thank you!!
Unfortunately, it is not possible for us to trouble-shoot work that we cannot see. How did you get a strictly numerical fraction? What happened to the variable?

#### helpmepls!

##### New member
I Literally just dived by 3 on both sides and thought that was the answers. formulas are so confusing

#### HallsofIvy

##### Elite Member
Divided what by 3?

Oh, dear, I think I know what you are saying- and it isn't pretty!

You started with $$\displaystyle \left(\frac{w^3}{w^7}\right)^3$$ and "divided by 3" to get $$\displaystyle \frac{w^3}{w^7}$$, getting rid of that "3" outside the parentheses.

No, first, while you can divide on both sides of an equation and still have a true equation, you do not have an equation here- you have an "expression" and doing any new arithmetic operation to it will change the result. Second, That "3" is an exponent, not a multiplier, so "divide by 3" does not get rid of it.

And once you had (incorrectly) reduced to $$\displaystyle \frac{w^3}{w^7}$$ you "cancel" the "w"s to get $$\displaystyle \frac{3}{7}$$. Again that is wrong. You cannot just cancel bases of exponents like that.