# 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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@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 Jul 14 '12 at 21:22
Thanks a lot. It helped tremendously! –  Austin Broussard Jul 14 '12 at 22:38