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Write each expression in the form $ca^pb^q$

c) $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$ \begin{align*} &= \frac{a\left(\frac{2}{b}\right)}{1}*\frac{\left(\frac{a}{3}\right)}{3}=\dfrac{a^2\left(\frac{2}{b}\right)}{3}=\frac{a^2}{1}*\frac{2}{b}*\frac{1}{3}=\frac{2a^2}{3b}*\frac{b}{1}=\frac{2a^2b}{3}=\frac{2}{3}a^2b^1 \end{align*}

e) $\dfrac{a^{-1}}{(b^{-1})\sqrt{a}}$
\begin{align*} &= \frac{1}{(b^{-1})a\sqrt{a}}=\frac{1b}{1a^1a^{\frac{1}{2}}}=\frac{1b^1}{1a^{\frac{2}{3}}}=1a^{\frac{-2}{3}}b^1 \end{align*}

These are my steps. Any corrections help.

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Asking a question purely to "check answers" doesn't really work with the purpose of the site. I know it's tempting, but we're more about explanation rather than answer checking. Something like Yahoo Answers may be more appropriate for this. Though for the record, a), b), d) and f) are correct. –  Robert Mastragostino Jul 15 '12 at 3:02
    
@RobertMastragostino well, I knew I had a few wrong. But I didn't know which ones. So I asked to check to see which I had wrong then I would re-work them and ask for advice. I mean I don't know any other way to do that because I don't know which one's I exactly got wrong. But I knew I didn't have them all right. You see? –  Austin Broussard Jul 15 '12 at 3:04
    
A good way to check is just to plug in numbers. Try $a=2$ and $b=3$, plug both into a calculator and see if you get the same result both times. This also helps because you might see "oh, this is off by a factor of $2$. I didn't multiply by $a$ enough" and then trace back to where the mistake was made. –  Robert Mastragostino Jul 15 '12 at 3:08
    
Thanks a lot. I am editing c and e right now with my steps and ask for advice on what to do! Thanks for informing me. –  Austin Broussard Jul 15 '12 at 3:09
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3 Answers 3

up vote 1 down vote accepted

On what grounds did you move $b$ to the top on (c)? It's incorrect. You cannot just multiply by $\frac{b}{1}$ because it pleases you to do so. And, after you suddenly create a factor of $\frac{b}{1}$ ex nihilo, it would have cancelled with the denominator. So both the penultimate and antepenultimate equality signs are incorrect.

$$\begin{align*} \frac{a(\frac{2}{b})}{\frac{3}{a}} &= \frac{2a}{b}\frac{a}{3}\\ &= \frac{2}{3}\frac{a^2}{b}\\ &= \frac{2}{3}a^2b^{-1}. \end{align*}$$

(d) is almost correct, except that $1+\frac{1}{2}=\frac{3}{2}$, not $\frac{2}{3}$. So the exponent of $a$ is incorrect.

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I understand what you mean! Thanks a lot. –  Austin Broussard Jul 15 '12 at 3:32
    
@AustinBroussard: There was a small mistake in (d) as well, in the epxonent of $a$ –  Arturo Magidin Jul 15 '12 at 3:44
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I'll do the first in two ways, $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$, and we'll see what we think. First, I'm going to separate all the 'numbers'.

  1. $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}= \dfrac{2a\frac{1}{b}}{3\frac{1}{a}} = \frac{2}{3}\dfrac{a\frac{1}{b}}{\frac{1}{a}}$

    Now I'll do the $a$ terms.

    $\frac{2}{3}\dfrac{a \frac{1}{b}}{\frac{1}{a}} = \frac{2}{3} a \frac{1}{b} \cdot \frac{a}{1} = \frac{2}{3}a^2 \frac{1}{b}$.

    Finally, we know that $\frac{1}{b} = b^{-1}$. So we have $\frac{2}{3} a^2 b^{-1}$.

  2. Let's do it a different way. We can get rid of the $\frac{3}{a}$ on the bottom, as dividing by a fraction is the same as multiplying by its reciprocal.

    $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}} = a\left(\frac{2}{b}\right) \cdot \frac{a}{3} = a \frac{2a}{3b} = 2a^2 \frac{1}{3b} = \frac{2}{3} a^2 b^{-1}$.

And if I had any doubt, I could check my answer by plugging in some numbers and making sure that both sides give me the same number.

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I see. The first way is easier for me. Thank you a lot! –  Austin Broussard Jul 15 '12 at 3:31
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If you remember that dividing by a fraction is the same as multiplying by its inverse, we get at once:

$$c)\,\,\frac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}=\frac{2a}{b}\frac{a}{3}=\frac{2a^2}{3b}=\frac{2}{3}a^2b^{-1}$$

$$(e)\,\,\frac{a^{-1}}{b^{-1}a^{1/2}}=a^{-1-1/2}\,b=a^{-3/2}\,b$$

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