Number Unit

In the last section we looked at some expressions like, “What is the third root of twenty-seven, squared?”  The math is kind of ugly looking.

$\sqrt[3]{{27}^2}$ 

The procedures are clunky and it is very easy to lose sight of the objective.  What this expression is asking is what number cubed is twenty-seven squared.  You could always square the 27, to arrive at 729 and see if that is a perfect cube. 

There is a much more elegant way to go about this type of calculation.  Turns out if we rewrite this expression with a rational exponent, life gets easier.

$\sqrt[3]{{27}^2}={27}^{2/3}$

These two statements are the same.  They ask the same question, what number cubed is twenty-seven squared?

By now you should be familiar with perfect cubes and squares.  Hopefully you’re also familiar with higher powers of 2 and 3, as well as a few others.  For example, you should recognize that 625 is $5^4.$ If you don’t know that yet, a cheat sheet might be helpful.

Let’s look at our expression again.  If you notice that 27 is a perfect cube, then you can rewrite it like this:

${27}^{2/3}\to {\left(3^3\right)}^{2/3}$

Maybe you see what’s going to happen next, but if not, we have a power raised to another here, we can multiply those exponents.  Three times two-thirds is two.  This becomes three squared.

${\left(3^3\right)}^{2/3}\to 3^2=9$

Not too bad!  We factor, writing the base of twenty-seven as an exponent with a power that matches the denominator of the other exponent, multiply, reduce, done! 

 

 

Let’s look at another. 

Simplify:

${625}^{3/4}$ 

We mentioned earlier that 625 was a power of 5, the fourth power of five.  That’s the key to making these simple.  Let’s rewrite 625 as a power of five.

${\left(5^4\right)}^{3/4}$

We can multiply those exponents, giving us five-cubed, or 125.  Much cleaner than finding the fourth root of six hundred and twenty-five cubed.

What about something that doesn’t work out so, well, pretty?  Something where the base cannot be rewritten as an exponent that matches the denominator?

${32}^{3/4}$

This is where proficiency and familiarity with powers of two comes to play.  Thirty-two is a power of two, just not the fourth power, but the fifth.

${\left(2^5\right)}^{3/4}$

If we multiplied these exponents together we end up with something that isn’t so pretty, $2^{15/4}.$ We could rewrite this by simplifying the exponent, but there’s a better way.  Consider the following, and note that we broke the five twos into a group of four and another group of one. 

${\left(2^5\right)}^{3/4}={\left(2^1\cdot 2^4\right)}^{3/4}$

Now we’d have to multiply the exponents inside the parenthesis by $\mathrm{¾}$, and will arrive at:

$2^{3/4}\cdot 2^3$

Notice that $2^{3/4}$ is irrational, so not much we can do with it, but two cubed is eight.  Let’s write the rational number first, and rewrite that irrational number as a radical expression:

$8\sqrt[4]{2^3},or8\sqrt[4]{8}$.

There’s an even easier way to think about these rational exponents.  I’d like to introduce something called Logarithmic Counting.  For those who don’t know what logarithms are, that might sound scary.

Do you remember learning how to multiply by 5s…how you’d skip count?  (5, 10, 15, 20, …)  Logarithmic counting is the same way, except with exponents.  For example, by 2:  2, 4, 8, 16, 32, …  Well, what’s the fourth step of 2 when logarithmically counting?  It’s 16, right?  $$ 

Let’s look at ${16}^{3/4}$.  See the denominator of four?  That means we’re looking for a fourth root, a number times itself four times that equals 16.  The three, in the numerator, it says, what number is three of the four steps on the way to sixteen?

2     4     8    16

Above is how we get to sixteen by multiplying a number by itself four times.  Do you see the third step is eight?

Let’s see how our procedure looks:

Procedure 1: 

${16}^{3/4}={\left(2^4\right)}^{3/4}$ 

${\left(2^4\right)}^{3/4}=2^{\frac{4}{1}\times \frac{3}{4}}=2^3,or8.$

Procedure 2:

${16}^{3/4}=\sqrt[4]{{16}^3}$

$\sqrt[4]{{16}^3}=\sqrt[4]{{\left(2^4\right)}^3}$

$\sqrt[4]{{\left(2^4\right)}^3}=\sqrt[4]{2^4}\times \sqrt[4]{2^4}\times \sqrt[4]{2^4}$ 

$\sqrt[4]{2^4}\times \sqrt[4]{2^4}\times \sqrt[4]{2^4}=2\times 2\times 2,or8.$

The most elegant way is to realize the 16 is the fourth power of 2, and the fraction $\mathrm{¾}$ is asking us for the third entry. What is 3/4s of the way to 16 when multiplying (exponents)?

Let’s look at ${625}^{2/3}.$ Let’s do this three ways, first with radical notation, then by evaluating the base and simplifying the exponents, and then by thinking about what is two thirds of the way to 625.

Now this is going to be a tricky problem because 625 is NOT a perfect cube.  It is the fourth power of 5, though, which means that 125 (which is five-cubed) times five is 625.

Radical Notation: 

${625}^{2/3}=\sqrt[3]{{625}^2}$ 

$\sqrt[3]{{625}^2}=\sqrt[3]{{\left(5^4\right)}^2}\to \sqrt[3]{5^8}$

$\sqrt[3]{5^8}=\sqrt[3]{5^3\times 5^3\times 5^2}$

$\sqrt[3]{5^3\times 5^3\times 5^2}=5\times 5\sqrt[3]{25},or25\sqrt[3]{25}$

Pretty ugly!

Exponential Notation:

${625}^{2/3}={\left(5^4\right)}^{2/3}$

${\left(5^4\right)}^{2/3}={\left(5^3\times 5^1\right)}^{2/3}$

${\left(5^3\times 5^1\right)}^{2/3}=5^2\times 5^{2/3}$

$5^2\times 5^{2/3}=25\times 5^{2/3},or25\sqrt[3]{25}$

A little better, but still a few sticky points.

Now our third method.

${625}^{2/3}$ asks, “What is two thirds of the way to 625, for a cubed number?” 

This 625 isn’t cubed, but a factor of it is. 

${625}^{2/3}={\left(125\times 5\right)}^{2/3}$
This could also be written as:

${125}^{2/3}\times 5^{2/3}$

I am certain that 5 to the two-thirds power is irrational because, well, five is a prime number.  Let’s deal with the other portion.

The steps to 125 are:  5    25     125

The second step is 25.

${125}^{2/3}\times 5^{2/3}=25\times 5^{2/3}$

To summarize the denominator of the rational exponent is the index of a radical expression.  The numerator is an exponent for the base.  How you tackle the expressions is entirely up to you, but I would suggest proficiency in multiple methods as sometimes the math lends itself nicely to one method but not another.

 

 

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Practice Problems

 

 

 

$$ $\begin{array}{l}\text{Simplify the following:}\\ \text{1}\text{. }{\left(16x^{16}\right)}^{3/4}\\ \\ \\ 2.{128}^{5/6}\\ \\ \\ \\ 3.\sqrt[5]{{125}^3}\\ \\ \\ \\ 4.{32}^{3/5}\\ \\ \\ \\ 5.{\left(81x^{27}\right)}^{2/3}\end{array}$ 

 

 

 

 

 

 

 

 

 

 

Select Practice Problems Review

 

1.  ${\left(16x^{16}\right)}^{3/4}$  A key piece of information that can make this easy is that the base, 16, is a perfect 4^th power.  Four is the denominator of the exponent.  So what we will do is rewrite 2⁴ in place of 16.

 

${\left(16x^{16}\right)}^{3/4}={\left(2^4{x}^{16}\right)}^{3/4}$

Inside the parenthesis are two bases, the 2 and the x.  Both bases get their exponents multiplied by $\mathrm{¾}$. 

${\left(2^4{x}^{16}\right)}^{3/4}=2^3{x}^{12}$

 

3.  $\sqrt[5]{{125}^3}$ This is a tricky problem because 125 is not a 5^th power!  The first thing I’d suggest is rewriting the radical notation as an exponent, and then writing 125 as 5³.

$\sqrt[5]{{125}^3}={\left(5^3\right)}^{3/5}$

Here we need to be delicate.  When we multiply 3 and 3/5, we get 9/5.  This is 1 and 4/5.

${\left(5^3\right)}^{3/5}=5^{9/5}\to 5\times 5^{4/5}$

You could leave the answer as 5 × 5^4/5, or rewrite it in radical notation. 

$\sqrt[5]{{125}^3}=5\sqrt[5]{5^4}\text{}\text{ or}5\times 5^{4/5}$