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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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    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.2012-07-15

3 Answers 3

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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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    @AustinBroussard: There was$a$small mistake in (d) as well, in the epxonent of $a$2012-07-15
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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!2012-07-15
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If you remember that dividing by a fraction is the same as multiplying by its inverse, we get at once:

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

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