# Problems regarding exponents

Write each of the following expressions in the form $ca^pb^q$ where $c$, $p$, $q$ are numbers:

1. $\dfrac{(2a^2)^3}{b}$ solved

2. $\sqrt{9ab^3}$ solved

3. $\dfrac{a(2/b)}{3/a}$ solved

4. $\dfrac{ab-a}{b^2-b}$ I tried and got to, $(ab-a)(b^2-b)^{-1}$. I know I'm supposed to bring $b^2-b$ to the top somehow because the answer calls for no fractions. That's all I have for that one.

5. $\dfrac{a^{-1}}{b^{-1}\sqrt{a}}$ I've figured out that $\sqrt(a) = a^{\frac{1}{2}}$. I also brought $b$ to the top and $a$ to the bottom to acquire; $1b^1/1a^1(a^{\frac{1}{2}})$. That's as far as I've gotten on that problem.

6. $\left(\dfrac{a^{2/3}}{b^{1/2}}\right)^2 \cdot \dfrac{b^{3/2}}{a^{1/2}}$ I am completely clueless on this one. Any help would be accepted.

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Dear Austin Broussard: "Welcome to math.SE. Since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the [homework] tag; people will still help, so don't worry." (quoted text by Arturo Magidin.) –  user2468 Jul 14 '12 at 20:09
@J.D. Arturo got it from me, though I don't think there's a need to quote anyone for it. –  Zev Chonoles Jul 14 '12 at 20:11

For your first one, $(a^m)^n=a^{mn}$ and $\frac 1 {a^n}=a^{-n}$

For your second, $\sqrt{a}=a^{1/2}$

Show us some attempts at solutions using those. I almost guarantee that you won't get help if you don't explain what you've tried. Partly because it's harder to explain a solution if we don't know where your understanding is lacking, and partly because we really don't want to become some kind of homework database.

EDIT: fantastic! Now we can get somewhere:

4: you can factor the numerator and denominator. Do that first.

5: this one's really gross. Write $a^{-1}$ and $b^{-1}$ as $\frac 1 a$ and $\frac 1 b$ and simplify the resulting fraction. then apply exponent rules. Namely $a^ma^n=a^{m+n}$.

6: $\dfrac{b^{3/2}}{a^{1/2}}=\dfrac{a^{-1/2}}{b^{-3/2}}$. Writing it like this (i.e. introducing negatives on purpose) makes the numerators and denominators play nice together. Then use $(a^m)^n=a^{mn}$ and $a^ma^n=a^{m+n}$.

Welcome to math.SE!

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Thank you for informing me. I'm obviously new to this site. –  Austin Broussard Jul 14 '12 at 22:38
@AustinBroussard wonderful! I've edited with more information. –  Robert Mastragostino Jul 14 '12 at 23:00

For the fourth one: $ab-a = a(b-1)$ and $b^2-b = b(b-1)$, so $\frac{ab-a}{b^2-b} = \frac{a(b-1)}{b(b-1)}$.

For the fifth one: Setting $\sqrt{a} = a^{1/2}$ and bringing $b$ to the top is good. You don't have to put $a$ in the denominator, however. Remember that $\frac1{a} = a^{-1}$, and that $(a^m)^n = a^{mn}$. You can use those to put together all the $a$'s.

For the sixth one: First, distribute the square over the fraction and remember, as I just said, that $(a^m)^n = a^{mn}$. You can use that to transform $\displaystyle \left(\frac{a^{2/3}}{b^{1/2}}\right)^2$ into something that looks like $\frac{a^x}{b^y}$. Then use the fact that, just as $a^{m+n} = a^m a^n$, $\frac{a^m}{a^n} = a^{m-n}$.

If you need more help, let me know.

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You might want to use different variables for the exponents, since the OP already uses $b$ for a different purpose. –  Rick Decker Jul 14 '12 at 21:21
@Rick: Yeah, I think I'll change it. Thanks for the advice! –  Javier Badia Jul 14 '12 at 21:22
Thanks a lot. It helped tremendously! –  Austin Broussard Jul 14 '12 at 22:38