Fractional Exponents Lesson

In this lesson, we will look at the rules pertaining to having a rational number as an exponent. This is also referred to as having fractional exponents. At this point, we know the product rule for exponents. This tells us if we multiply two expressions in exponent form with the same base, we keep the base the same and add exponents. Let's suppose we saw: $$x^3 \cdot x^5$$ Since the base (x) is the same in each case, keep the base (x) the same and add exponents (3 + 5): $$x^3 \cdot x^5=x^{(3 + 5)}=x^8$$ So what happens if we see something such as: $$x^{\frac{1}{2}}\cdot x^{\frac{1}{2}}$$ If we apply the product rule for exponents: $$x^{\frac{1}{2}}\cdot x^{\frac{1}{2}}=x^{\left(\frac{1}{2}+ \frac{1}{2}\right)}=x^1$$ This suggests that: $$x^{\frac{1}{2}}=\sqrt{x}$$ Since: $$\sqrt{x}\cdot \sqrt{x}=x$$ Similarly, we could look at: $$x^{\frac{1}{3}}\cdot x^{\frac{1}{3}}\cdot x^{\frac{1}{3}}$$ $$x^{\frac{1}{3}}\cdot x^{\frac{1}{3}}\cdot x^{\frac{1}{3}}=x^{\left(\frac{1}{3}+ \frac{1}{3}+ \frac{1}{3}\right)}=x^1$$ This suggests that: $$x^{\frac{1}{3}}=\sqrt[3]{x}$$ Since: $$\sqrt[3]{x}\cdot \sqrt[3]{x}\cdot \sqrt[3]{x}=x$$ Let's define our first rule for fractional exponents:
let a ≥ 0
let n > 0
$$a^{\frac{1}{n}}=\sqrt[n]{a}$$ Let's look at a few examples.
Example 1: Simplify each. $$32^{\frac{1}{5}}$$ This is telling us we have the fifth root of 32. The base (32) of the exponential expression becomes the radicand (number under the radical symbol). The denominator (5) of the fraction part becomes our index for the radical. $$32^{\frac{1}{5}}=\sqrt[5]{32}$$ What number multiplied by itself five times gives us 32? The answer is 2:
32 = 2⁵ $$32^{\frac{1}{5}}=\sqrt[5]{32}=2$$ Example 2: Simplify each. $$216^{\frac{1}{3}}$$ This is telling us we have the third (cube) root of 216. The base (216) of the exponential expression becomes the radicand (number under the radical symbol). The denominator (3) of the fraction part becomes our index for the radical. $$216^{\frac{1}{3}}=\sqrt[3]{216}$$ What number multiplied by itself three times gives us 216? The answer is 6:
216 = 6³ $$216^{\frac{1}{3}}=\sqrt[3]{216}=6$$ Now that we have the first rule down, let's think about another scenario. Suppose we saw the following: $$4^{\frac{3}{2}}$$ Using the power to power rule for exponents, we could write this problem as: $$4^{\frac{3}{2}}=(4^{\frac{1}{2}})^3$$ At this point, we know that raising a number to the 1/2 power is the same as taking the square root of the number: $$(4^{\frac{1}{2}})^3=(\sqrt{4})^3=2^3=8$$ Additionally, we could have also done this in a different order: $$4^{\frac{3}{2}}=(4^3)^{\frac{1}{2}}$$ In this case, we are raising 4 to the third power first. This will give us 64: $$(4^3)^{\frac{1}{2}}=64^{\frac{1}{2}}$$ Now we can take the square root of 64, which is 8. Notice how we get the same answer either way: $$64^{\frac{1}{2}}=\sqrt{64}=8$$ When working with this type of problem, it is normally suggested to take the root first. This will result in working with a smaller number. Since both methods produce the same result, we are free to work the problem either way.
Let's now define our second rule for fractional exponents:
let a ≥ 0
let m and n be integers
let n > 0 $$a^{\frac{m}{n}}=(\sqrt[n]{a})^m$$ $$a^{\frac{-m}{n}}=\frac{1}{(\sqrt[n]{a})^m}$$ Let's look at a few examples.
Example 3: Simplify each. $$125^{\frac{4}{3}}$$ From our rule, we can begin by taking the cube root of 125 and raising the result to the 4th power: $$125^{\frac{4}{3}}=(\sqrt[3]{125})^4$$ The cube root of 125 is 5. We can replace this in our problem: $$(\sqrt[3]{125})^4=5^4$$ Now we can raise 5 to the 4th power: $$5^4=625$$ $$125^{\frac{4}{3}}=625$$ Example 4: Simplify each. $$\large{\frac{y^{\frac{1}{2}}\cdot \left(y^{\frac{-5}{4}}\right)^{-2}}{x^{\frac{4}{3}}y^{\frac{-1}{3}}}}$$ Let's work on the numerator first: $$\large{y^{\frac{1}{2}}\cdot \left(y^{\frac{-5}{4}}\right)^{-2}=}$$ $$\large{y^{\frac{1}{2}}\cdot y^{\frac{(-5)(-2)}{4}}=}$$ $$\large{y^{\frac{1}{2}}\cdot y^{\frac{5}{2}}=}$$ $$\large{y^{\frac{1}{2}+ \frac{5}{2}}=}$$ $$\large{y^{\frac{6}{2}}=}$$ $$\large{y^{3}}$$ Let's replace the numerator in our problem: $$\large{\frac{y^3}{x^{\frac{4}{3}}y^{\frac{-1}{3}}}}$$ Let's now work on our denominator: x to the 4/3 power can't be simplified, but we can use our quotient rule for exponents to work with y to the power of -1/3: $$\large{\frac{y^{3}}{x^{\frac{4}{3}}y^{\frac{-1}{3}}}}=$$ $$\large{\frac{y^{3 - \left( \frac{-1}{3}\right)}}{x^{\frac{4}{3}}}}=$$ $$\large{\frac{y^{\frac{9}{3}+ \left( \frac{1}{3}\right)}}{x^{\frac{4}{3}}}}=$$ $$\large{\frac{y^{\frac{10}{3}}}{x^{\frac{4}{3}}}}$$ Since x to the 4/3 power technically involves a radical, we need to rationalize the denominator. Generally speaking, we should clear fractional exponents from any denominator so our result is simplified. To make this really easy, think about the denominator (3). What is the next highest value which would be divisible by 3? The answer is 6 since 6/3 is 2. This means we can multiply the numerator and denominator by x to the 2/3 power: $$\large{\frac{y^{\frac{10}{3}}}{x^{\frac{4}{3}}}}\cdot \frac{x^{\frac{2}{3}}}{x^{\frac{2}{3}}}=$$ $$\large{\frac{y^{\frac{10}{3}}x^{\frac{2}{3}}}{x^{\frac{2 + 4}{3}}}}=$$ $$\large{\frac{y^{\frac{10}{3}}x^{\frac{2}{3}}}{x^2}}$$

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