# Simplify $\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} * \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$ to $\frac{\sqrt{mnc}}{a^9cmn}$

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 $$\dfrac{\sqrt{mnc}}{a^9cmn}$$?

• $a^{-7}n^{-2}={1\over a^7n^2}$ Jan 8 '19 at 16:55

## 5 Answers

\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}

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

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

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

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}

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

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.)

• 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? Jan 8 '19 at 17:14
• 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. Jan 8 '19 at 17:16

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 $$\frac{\sqrt{mnc}}{\sqrt{mnc}}$$

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