# 5.6.1 Rational Exponents: 1/n

• x^{mn}=(x^m)^n=(x^n)^m
• x^{1/n}=\sqrt[n]{x} is the nth root of x.

The next piece of the puzzle is to try to understand $x^{m/n}$ where m and and n are counting numbers; this will help us to compute the population of our wildebeests at time 1.5 years, since 1.5 is just 3/2.

You may recall that we studied x^{m+n} before we looked at x^{m-n}; addition is oftentimes easier to understand than subtraction. Now, we're hoping to study x^{m/n}. However, we don't yet have a firm understanding of x^{mn}. Since division is the inverse process of multiplication and multiplication is much easier to understand than division, let's start with x^{mn} and see where it takes us.

As usual, let's start with a specific example. With m=3 and n=2, can we rewrite $x^{3\times2}$ as something else?

If we realize that 3 × 2 = 6 = 3 + 3, then we can see that

x^{3 \times 2} =x^6=x^3x^3=(x^3)^2

Here we used the fact that when multiplying exponentials, we can add exponents if we have the same base.

But, we can also think of 6 as being equal to 2 + 2 + 2.

With this in mind, we can see that

x^{3 \times 2} =x^6=x^2x^2x^2=(x^2)^3

This tells us that

$x^{3 \times 2} =(x^3)^2=(x^2)^3$

Since there was nothing special about 2, 3, and 6, let's jump up a level of abstraction, developing a new general property of exponents. From the work above, if m and n are counting numbers, we can write x^{mn} in two ways:

x^{mn}=(x^m)^n=(x^n)^m

It turns out that the same property holds for integers as well. Without going through a formal mathematical proof, let's just explore this concept through one example. If m=2 and n=-3, then we have that:

\begin{align*}\left ( x^{-3} \right )^2&=\left ( \frac{1}{x^3} \right )^2\\&=\frac{1}{x^3}\cdot\frac{1}{x^3}\\&=x^{-3}x^{-3}\\&=x^{-6}\end{align*}

We then have that, if m and n are integers,

x^{mn}=(x^m)^n=(x^n)^m

Explore!

Evaluate the following without a calculator:

(2^3)^2=

(3^{-2})^{-2}

Now that we now have a deeper understanding of x to "products of integers", let's go back to our exponential of interest, namely x^{m/n}.

If we're to treat this expression in a way that follows the same pattern as above, then since

\frac{m}{n}=m \times \frac{1}{n}=\frac{1}{n}\times m,

we have that

x^{m/n} = x^{m\times\left (1/n\right)} = \left (x^m\right)^{1/n} = \left (x^{1/n}\right)^m

Let's use specific numbers for m and n to understand what the fractional exponents 1/n and 1/m actually mean. Using the property above, we see that

(4^{1/2})^2=

:

Even though we don't know what 4^{1/2} means yet, we now see that if we square it, we get back 4. Any thoughts about what the symbol could represent?

Let's try another one:

(8^{1/3})^3=

:

Even though we don't know what 8^{1/3} means, we see that if we cube it, we get back 8.

Putting the two examples together, it looks like raising something to the 1/2 power is the same thing as taking the square root; raising it to the 1/3 power must be the same as taking the cube root!

In general, then, we have that

x^{1/n} is the nth root of x. We also write the nth root as \sqrt[n]{x}.

Explore!

Compute the following without using a calculator:

64^{1/2}=

125^{1/3}=

With the tools from this section, we're ready to understand what it would mean to put 1.5 into our original model:

P(1.5)=P(3/2)=1000(1.02)^{3/2}

We'll finish this example in the next section.

My Book

Key Points

Extra Practice
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