1
$\begingroup$

I need to simplify $$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$$

The solution provided is: $\dfrac{\sqrt{mnc}}{a^9cmn}$.

I'm finding this challenging. I was able to make some changes but I don't know if they are on the right step or not:

First, I am able to simplify the left fractions numerator and the right fractions denominator:

$\sqrt{mn^3}=\sqrt{mn^2n^1}=n\sqrt{mn}$

$\sqrt{m^2c^4}=m\sqrt{c^2c^2} = mcc$

So my new expression looks like:

$$\frac{n\sqrt{mn}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{mcc}$$

From this point I'm really at a loss to my next steps. If I multiply them both together I get:

$$\frac{(n\sqrt{mn})(a^{-7}n^{-2})}{(a^2\sqrt{c^{-3}})(mcc)}.$$

Next, I was thinking I could multiply out the radical in the denominator but I feel like I need to simplify what I have before going forwards.

Am I on the right track? How can I simplify my fraction above in baby steps? I'm particularly confused by the negative exponents.

How can I arrive at the solution $\dfrac{\sqrt{mnc}}{a^9cmn}$?

$\endgroup$
  • 2
    $\begingroup$ $a^{-7}n^{-2}={1\over a^7n^2}$ $\endgroup$ – saulspatz Jan 8 at 16:55
2
$\begingroup$

$$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}}\cdot\frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}} = \frac{\sqrt{mn^3}a^{-7}n^{-2}}{a^2\sqrt{m^2c^{-3}c^4}}$$

From here, you use the following identity:

$$a^{-b} = \frac{1}{a^b}$$

You can simplify from here:

$$= \frac{\sqrt{mn^3}}{a^9n^2\sqrt{m^2c}} = \frac{\sqrt{n^3}}{a^9n^2\sqrt{mc}} = \color{blue}{\frac{\vert n\vert\sqrt{n}} {a^9n^2\sqrt{mc}}} = \color{green}{\frac{\sqrt{nmc}}{a^9cm\sqrt{n^2}}} = \frac{\sqrt{nmc}}{a^9cm\vert n\vert}$$

Notice the step highlighted in blue. Clearly, when dealing with real numbers, the radicand must be non-negative (and the denominator can’t be $0$), so $mc > 0$. Therefore, in the next step, we can note that $nmc > 0$, and since $mc > 0$, then $n > 0$, so the absolute value of $n$ is $n$ itself. (Mathematically, $\vert n\vert = n$.) Hence, the final result becomes

$$\frac{\sqrt{nmc}}{a^9cmn}$$

Addition: This is, of course, not to make the problem seem more confusing than it actually is. However, it is a common error to forget the absolute value sign when dealing with even indices. $\sqrt{a^2}$ is not $a$, it’s $\vert a\vert$ because $a$ itself may be negative, but the returned value is always non-negative. For example, $\sqrt{(-2)^2} = \sqrt{4} = 2 = \vert -2\vert$. (This also applies to all even indices, such as fourth roots, sixth roots, etc. Always be careful when dealing with these.)

$\endgroup$
  • $\begingroup$ Hi KM, after you write "You can simplify from here:" you have a row of 5 sequential fractions. Moving from the first to the second you removed m in the numerator. How did you do that? $\endgroup$ – Doug Fir Jan 8 at 17:14
  • 1
    $\begingroup$ You have $\sqrt{m}$ in the numerator and $\sqrt{m^2}$ in the denominator, so you can divide them, leaving you with $\sqrt{m}$ in the denominator. $\endgroup$ – KM101 Jan 8 at 17:16
4
$\begingroup$

\begin{align} \frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}} &= \frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{1}{a^7n^{2} mc^2}\\ & = \frac{\sqrt{mn^3}}{a^2} \cdot \frac{\sqrt{c^{3}}}{a^7n^{2}mc^2} \\ & = \frac{\sqrt{m}n\sqrt{n}c\sqrt{c}}{a^9 n^{2}m c^2} \\ & = \frac{\sqrt{nmc}}{a^9 n m c} \\ \end{align}

$\endgroup$
2
$\begingroup$

$\cfrac{\sqrt{mn^{3}}}{a^{2}\sqrt {c^{-3}}} = \cfrac{\sqrt{mn}\lvert n\rvert}{a^{2}\sqrt {c^{-3}}} = \cfrac{\sqrt{mnc}\lvert n\rvert}{a^{2}\sqrt {c^{-2}}} = \cfrac{\sqrt{mnc}\lvert n\rvert \lvert c\rvert}{a^{2}}$

$\cfrac{\sqrt{mnc}\lvert n\rvert \lvert c\rvert}{a^{2}} * \cfrac{a^{-7}n^{-2}}{c^{2}\lvert m\rvert} = \cfrac{\sqrt{mnc}\lvert n\rvert \lvert c\rvert}{a^{9}n^{2}c^{2}m} = \cfrac{\sqrt{mnc}}{a^{9}cmn} $

Oof, took a while to edit. The last equality holds when $\lvert nc\rvert = nc$

$\endgroup$
2
$\begingroup$

Use $$a^{-b}=\frac{1}{a^{b}} .$$

So \begin{align} \frac{(n\sqrt{mn})(a^{-7}n^{-2})}{(a^2\sqrt{c^{-3}})(mc^2)}&=\frac{n\sqrt{mn}}{(a^2{c^{-3/2}})(mc^2)(a^{7}n^{2})}\\ &=\frac{(\sqrt{mn})c^{3/2}}{(mc^2)(a^{9}n)}\\ &=\frac{\sqrt{mnc}}{a^{9}cmn}\\ \end{align}

$\endgroup$
1
$\begingroup$

Recall that $a^{-x}=\frac{1}{a^x}$, $a^{1/x}=\sqrt[x]{a}$, and $a^na^m=a^{m+n}$

Always convert everything to exponents first, then use arithmetic

$$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}}\frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}=m^{1/2}\ n^{3/2}\ a^{-2}\ c^{3/2}\ a^{-7}\ n^{-2}\ m^{-2/2}\ c^{-4/2}$$ Rearrange $$m^{1/2}\ n^{3/2}\ a^{-2}\ c^{3/2}\ a^{-7}\ n^{-2}\ m^{-2/2}\ c^{-4}=(m^{1/2}\ m^{-2/2})\ (n^{3/2}\ n^{-2})\ (a^{-2}\ a^{-7})\ (c^{3/2}\ c^{-4/2})$$ Simplify $$(m^{1/2-1})\ (n^{3/2-2})\ (a^{-2-7})\ (c^{3/2-2})=(m^{-1/2})\ (n^{-1/2})(a^{-9})(c^{-1/2})=\frac{1}{a^9\sqrt{mnc}}$$

Multiply by $\frac{\sqrt{mnc}}{\sqrt{mnc}}$

$$\frac{1}{a^9\sqrt{mnc}}\frac{\sqrt{mnc}}{\sqrt{mnc}}=\frac{\sqrt{mnc}}{a^9mnc}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.