Have you ever been in a science laboratory? Take a look at this dilemma.

One of the places that the students were able to visit when they went downtown was a laboratory at the city college. Downtown, the city college had some of its classrooms and one of the classrooms was a laboratory.

“This is a good friend of mine Professor Smith,” Mr. Travis said introducing the students to a woman with blonde hair and a wide smile.

“Welcome,” Professor Smith said. “Are you enjoying your trip downtown?”

Many students responded yes and then were drawn over to one of the laboratory tables where a lot of work was taking place.

“What is happening here?” Sam asked.

“Well, I started with a very small sample of cobalt. I actually had 10 grams of it and I took a third of a third of a third of a third of it,” She explained.

The students began figuring the math out in their heads.

**
Can you figure it out? How many grams did the sample end up being? By the end of this Concept, you will be able to solve this dilemma.
**

### Guidance

This may sound confusing, but in math, we can rewrite this as \begin{align*}\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\end{align*} or @$\begin{align*}\left(\frac{1}{2} \right)^4\end{align*}@$ . We can use exponents with fractions or quotients, too. In order to answer the question above, we would multiply the numerators and denominators across, like this: @$\begin{align*}\frac{1 \cdot 1 \cdot 1 \cdot 1}{2 \cdot 2 \cdot 2 \cdot 2}=\frac{1}{16}\end{align*}@$ . Half of a half of a half of a half is one sixteenth. Once again, we have repeating multiplication of the same number which we could write more easily as @$\begin{align*}\frac{1^4}{2^4}=\frac{1}{16}\end{align*}@$ .

**
The
**
**
Power of a Quotient Property
**

**says that for any nonzero numbers @$\begin{align*}a\end{align*}@$ and @$\begin{align*}b\end{align*}@$ and any integer @$\begin{align*}n\end{align*}@$ :**

@$$\begin{align*}\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}\end{align*}@$$

Here is one for you to try.

@$\begin{align*}\left(\frac{5}{3}\right)^4=\frac{5^4}{3^4}=\frac{625}{81}\end{align*}@$

**
You can see in this situation that we have simplified the expression by figuring out what five to the fourth is and what three to the fourth is. The next step in this problem would be to divide.
**

Take a look at this one.

@$\begin{align*}\left(\frac{3k}{2j}\right)^4=\frac{(3k)^4}{(2j)^4}=\frac{(3k)(3k)(3k)(3k)}{(2j)(2j)(2j)(2j)}=\frac{81k^4}{16j^4}\end{align*}@$

**
This problem has different variables, so this is as far as we can take this problem.
**

Simply each quotient.

#### Example A

@$\begin{align*}\left(\frac{4}{5}\right)^3 \end{align*}@$

**
Solution:
@$\begin{align*}\frac{64}{125} = .512\end{align*}@$
**

#### Example B

@$\begin{align*}\left(\frac{2a}{3b}\right)^2 \end{align*}@$

**
Solution:
@$\begin{align*}\frac{4a^2}{9b^2}\end{align*}@$
**

#### Example C

@$\begin{align*}\left(\frac{a}{5b}\right)^3 \end{align*}@$

**
Solution:
@$\begin{align*}\frac{a^3}{125b^3}\end{align*}@$
**

Now let's go back to the dilemma from the beginning of the Concept.

**
To figure out the number of grams in the sample, we must use what we have learned about monomials and powers.
**

**
Professor Smith started off with 10 grams.
**

**
Then she took a third of a third of a third of a third of it. That is
@$\begin{align*}\frac{1}{3}\end{align*}@$
to the fourth power.
**

**
Here is how we can set up the problem.
**

@$\begin{align*}10 \left(\frac{1}{3}\right)^4=10 \left(\frac{1^4}{3^4}\right)=10 \left (\frac{1}{81}\right)=\frac{10}{81} \ grams\end{align*}@$

**
We can convert that into a decimal by dividing the numerator by the denominator.
**

**
.12 grams is our answer as a decimal.
**

### Guided Practice

Here is one for you to try on your own.

Simplify the following quotient.

@$\begin{align*}\left(\frac{-4x}{3y}\right)^3 \end{align*}@$

**
Solution
**

First, let's work with the numerator.

@$\begin{align*}(-4x)^3 = -64x^3\end{align*}@$

Now let's work with the denominator.

@$\begin{align*}(3y)^3 = 27y^3\end{align*}@$

Here is our final answer.

@$\begin{align*}\frac{-64x^3}{27y^3}\end{align*}@$

### Video Review

Multiplying and Dividing Monomials

### Explore More

Directions: Simplify.

- @$\begin{align*}\left(\frac{2}{3}\right)^4 \end{align*}@$
- @$\begin{align*}\left(\frac{1}{3}\right)^3 \end{align*}@$
- @$\begin{align*}\left(\frac{7}{8}\right)^2 \end{align*}@$
- @$\begin{align*}\left(\frac{2}{5}\right)^4 \end{align*}@$
- @$\begin{align*}\left(\frac{7k}{-2m}\right)^3\end{align*}@$
- @$\begin{align*}\left(\frac{3x}{-2y}\right)^3\end{align*}@$
- @$\begin{align*}\left(\frac{4x}{-3y}\right)^4\end{align*}@$
- @$\begin{align*}\left(\frac{5y}{-2z}\right)^5\end{align*}@$
- @$\begin{align*}\left(\frac{-2y}{4z}\right)^4\end{align*}@$
- @$\begin{align*}\left(\frac{4xy}{-2z^5}\right)^5\end{align*}@$
- @$\begin{align*}\left(\frac{12x^2y^4}{-6z^3}\right)^2\end{align*}@$
- @$\begin{align*}\left(\frac{7x^2y}{-2z^3}\right)^3\end{align*}@$
- @$\begin{align*}\left(\frac{2x^3y^2}{-2z^3}\right)^3\end{align*}@$
- @$\begin{align*}\left(\frac{x^{11}}{y^9}\right)^5\end{align*}@$
- @$\begin{align*}\left(\frac{-5x^3}{3h^2 j^8}\right)^5\end{align*}@$