Indices

UK Secondary (7-11)

Fractional Indices with Algebraic Terms

Lesson

The fractional index rule states:

$x^{\frac{1}{n}}=\sqrt[n]{x}$`x`1`n`=^{n}√`x`

Let's look at *why* this is true using the multiplication law.

Let's say I wanted to solve $25^{\frac{1}{2}}\times25^{\frac{1}{2}}$2512×2512. Using the multiplication law, this would be the same as $25^{\frac{1}{2}+\frac{1}{2}}=25^1$2512+12=251 or $25$25. When we multiply a number by itself, the end product is called a *square number*. Since we know that $25^{\frac{1}{2}}\times25^{\frac{1}{2}}=25$2512×2512=25, $25^{\frac{1}{2}}$2512 must equal the *square root* of $25$25, which is $5$5. This must be right as $5\times5=25$5×5=25.

You may also see questions with more complicated fractional indices, such as $x^{\frac{3}{2}}$`x`32. We could express this as a power of a power, $\left(x^3\right)^{\frac{1}{2}}$(`x`3)12. As such, the **numerator** in the fractional index can be expressed as a **power** and the **denominator** in the fractional index can be expressed as a **root**.

More generally, this rule states:

$x^{\frac{m}{n}}=\sqrt[n]{x^m}$`x``m``n`=^{n}√`x``m`

When solving problems with fractional indices, it doesn't matter whether you start with the powers or the roots (although you might find it easier to do it one way than the other). For example, let's look at $16^{\frac{3}{2}}$1632:

$\sqrt{16^3}=4^3$√163=43

$4^3=64$43=64

Now let's start with the powers:

$\sqrt{16^3}=\sqrt{4096}$√163=√4096

$\sqrt{4096}=64$√4096=64

See, we get the same answer both ways.

**Evaluate:** $121^{\frac{1}{2}}$12112.

**Think:** A fractional index of $\frac{1}{2}$12 is the same as finding the square root.

**Do:**

$121^{\frac{1}{2}}$12112 | $=$= | $\sqrt{121}$√121 |

$=$= | $11$11 |

Express as a fraction in simplest form.

$625^{\frac{-3}{4}}$625−34

Express $\sqrt[5]{x^7}$^{5}√`x`7 in index form.