I work as an after school tutor at my high school. I've had kids come up to me asking how to do these types of problems:

$\left(\displaystyle \frac{5xy^{-2}}{3z^{-1}} \right)^{-2}$

My approach is to tell them to use the power of quotient rule:

$\left(\displaystyle \frac{a}{b} \right)^{n} = \displaystyle \frac{a^n}{b^n}$

$\displaystyle \frac{\left(5xy^{-2}\right)^{-2}} {\left(3z^{-1} \right)^{-2}}$

Then I tell them to use power of a product: $(a\cdot b)^{n} = a^{n}b^{n}$

$\displaystyle \frac{5^{-2} x^{-2} \left(y^{-2}\right)^{-2}}{3^{-2}\left(z^{-1} \right)^{-2}}$

Freshmen find these steps confusing and when they get to this step they don't even know what to do, but I thought these rules would be helpful. I even gave them examples of each rule. The tutoring hours after school are usually filled with people struggling in algebra. How could I explain it better?


You may do it as follows:

$$\left(\displaystyle \frac{5xy^{-2}}{3z^{-1}} \right)^{-2} =\left(\frac{3z^{-1}}{5xy^{-2}}\right)^2=\left(\frac{3y^2}{5xz}\right)^2=\left(\frac{9y^4}{25x^2z^2}\right)$$ Note that $$\Box^{~\color{blue}{-\star}}=\frac{1}{\Box^{~\color{red}{+\star}}}$$


You could start off by showing them this...Using the rule $$\frac{a^n}{a^m}=a^{n-m}$$ $($I'm assuming that they are okay with the rule associated with dividing expressions with the same base$)$: $$a^3\div a^5=a^{3-5}=a^{-2}\space \text{and}\space a^3\div a^5=\frac{a\times a\times a}{a\times a\times a\times a\times a}=\frac{1}{a^2}\\ \text{so clearly}\\ a^{-2}=\frac{1}{a^2}$$ You can do some more examples, maybe two or three to give them some more intuition. Then you can introduce the general rule: $$a^{-n}=\frac{1}{a^n}$$

Now tackle the questions in steps.


Add brackets here so that they do not get confused and understand that only $y$ and $z$ are being raised to $-2$ and $-1$ in the numerator and denominator respectively. Now using the general rule for negative exponents from above, inside the brackets can be re-written and simplified as:

$$\left(\frac{5x(y^{-2})}{3(z^{-1})}\right)=\frac{5x\left(\frac{1}{y^2}\right)}{3\left(\frac{1}{z^1}\right)}=\frac{5x}{y^2} \div \frac{3}{z}=\frac{5x}{y^2}\cdot\frac{z}{3}=\frac{5xz}{3y^2}$$

Now you have the simplified expression of the inner brackets raised to $-2$. Apply the general rule again to get: $$\left(\frac{5xz}{3y^2}\right)^{-2}=\frac{1}{\left(\frac{5xz}{3y^2}\right)^{2}}$$

Now for the denominator..Give them some computations of the form $\frac{a^n}{b^n}$, $\left(\frac{a}{b}\right)^n$ and $a^nb^n$, $(ab)^n$ maybe three or four examples. After computing these, they should intuitively begin to realize that $\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n$ and $a^nb^n=(ab)^n$. After that you can explain to them that this is so $($i.e. they hold$)$ because of the power of quotient rule and power of a product rule respectively.

Applying this to the denominator it becomes: $$\left(\frac{5xz}{3y^2}\right)^{2}=\frac{(5xz)^2}{(3y^2)^2}=\frac{5^2x^2z^2}{3^2(y^2)^2}=\frac{5^2x^2z^2}{3^2y^4}$$

Here you can stress that when an expression is raised to a power, you need to raise everything in the expression to the power. And that is why $(3y^2)^2=3^2(y^2)^2=3^2y^{2(2)}=3^2y^4$

Alternately if they are still not following why $a^nb^n=(ab)^n$, another way to look at it could be like this. If you have $(ab)^2$ this is the same as multiplying "$ab$" by itself twice..If you have $(ab)^3$ this is the same as multiplying "$ab$" by itself thrice and so on and so on. So $$(5xz)^2=(5xz)(5xz)=25x^2z^2,\space {}\space (3y^2)^2=(3y^2)(3y^2)=9y^4$$

Simplifying the last bit becomes $$\frac{5^2x^2z^2}{3^2y^4}=\frac{25x^2z^2}{9y^4}$$

But remember that was just the denominator and it is being divided by $1$, so: $$\frac{1}{\frac{25x^2z^2}{9y^4}}=1\div \frac{25x^2z^2}{9y^4}=1\cdot\frac{9y^4}{25x^2z^2}=\frac{9y^4}{25x^2z^2}$$

  • 1
    $\begingroup$ Complete source and good points. +1 $\endgroup$ – Mikasa Jan 17 '14 at 16:58

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