The last index law looks at powers that are fractions:
We can define $x^{\frac{1}{n}}=\sqrt[n]{x}$x1n=n√x, that is $x^{\frac{1}{n}}$x1n is the $n^{th}$nth root of $x$x. It then follows from previous index laws that:
| $x^{\frac{m}{n}}$xmn | $=$= | $\left(x^{\frac{1}{n}}\right)^{^m}$(x1n)m |
| $=$= | $\left(\sqrt[n]{x}\right)^m$(n√x)m |
Or
| $x^{\frac{m}{n}}$xmn | $=$= | $\left(x^m\right)^{\frac{1}{n}}$(xm)1n |
| $=$= | $\sqrt[n]{x^m}$n√xm |
So in general, we have:
The definition of $x^{\frac{1}{n}}=\sqrt[n]{x}$x1n=n√x can be shown to be consistent with the previous index laws as follows:
| For $a>0$a>0: | $a^{\frac{1}{2}}\times a^{\frac{1}{2}}$a12×a12 | $=$= | $a^{\frac{1}{2}+\frac{1}{2}}$a12+12 |
| $=$= | $a^1$a1 | ||
| $=$= | $a$a | ||
| Just as: | $\sqrt{a}\times\sqrt{a}$√a×√a | $=$= | $a$a |
Similarly for higher roots:
| For $a>0$a>0: | $\left(a^{\frac{1}{n}}\right)^n$(a1n)n | $=$= | $a^{\frac{1}{n}\times n}$a1n×n |
| $=$= | $a^1$a1 | ||
| $=$= | $a$a | ||
| Just as: | $\left(\sqrt{n}\right)^n$(√n)n | $=$= | $a$a |
- For $a$a a positive real number then $a^{\frac{1}{n}}$a1n is the positive number whose $n^{th}$nth power is $a$a.
- For $a=0$a=0, we can define $0^{\frac{1}{n}}=0$01n=0 for any positive $n$n, since $0^n=0$0n=0.
- For $a$a a negative real number, even roots are not defined. However, for odd roots we can define it as the negative number whose $n^{th}$nth power is $a$a. For example $\left(-27\right)^{\frac{1}{3}}=-3$(−27)13=−3, since $\left(-3\right)^3=-27$(−3)3=−27.
When evaluating problems with fractional indices, it doesn't matter whether you start with the powers or the roots, although you might find it more efficient to do it one way than the other.
Worked examples
Example 1
Evaluate $16^{\frac{3}{2}}$1632.
We can start by taking the root then applying the power:
| $16^{\frac{3}{2}}$1632 | $=$= | $\left(\sqrt{16}\right)^3$(√16)3 |
| $=$= | $4^3$43 | |
| $=$= | $64$64 |
Or we could start by applying the power then taking the root.
| $16^{\frac{3}{2}}$1632 | $=$= | $\sqrt{16^3}$√163 |
| $=$= | $\sqrt{4096}$√4096 | |
| $=$= | $64$64 |
So, we get the same answer both ways. However, taking the root of $16$16 and then cubing $4$4 may be more efficient than dealing with the larger numbers.
Example 2
Simplify $\sqrt{27\times x^4}\div\sqrt[3]{3^6x^2}$√27×x4÷3√36x2.
Think: Write roots as fractional powers and simplify using our index laws. Notice we have a base of $27$27 and a base of $3$3, we can express the $27$27 as a base of $3$3 and then combine using our index laws.
Do:
| $\sqrt{27\times x^4}\div\sqrt[3]{3^6x^2}$√27×x4÷3√36x2 | $=$= | $\left(3^3\times x^4\right)^{\frac{1}{2}}\div\left(3^6x^2\right)^{\frac{1}{3}}$(33×x4)12÷(36x2)13 |
| $=$= | $3^{\left(3\times\frac{1}{2}\right)}x^{\left(4\times\frac{1}{2}\right)}\div3^{\left(6\times\frac{1}{3}\right)}x^{\left(2\times\frac{1}{3}\right)}$3(3×12)x(4×12)÷3(6×13)x(2×13) | |
| $=$= | $3^{\frac{3}{2}}x^2\div3^2x^{\frac{2}{3}}$332x2÷32x23 | |
| $=$= | $3^{\left(\frac{3}{2}-2\right)}x^{\left(2-\frac{2}{3}\right)}$3(32−2)x(2−23) | |
| $=$= | $3^{\frac{-1}{2}}x^{\frac{4}{3}}$3−12x43 | |
| $=$= | $\frac{x^{\frac{4}{3}}}{\sqrt{3}}$x43√3 |
Remember that when we raise a negative number to an even power, it becomes a positive number. For instance, $\left(-5\right)^2=25$(−5)2=25. This means that $\sqrt{(-5)^2}=\sqrt{25}=5$√(−5)2=√25=5.
If we now consider the algebraic expression $\sqrt{a^2}$√a2, the power of a power rule indicates that this should simplify to $a$a. As you can see above, however, this is not the case if $a$a is a negative number!
So be careful when simplifying even powers and roots of algebraic expressions - make sure to think about whether or not the variable could represent a negative number.
Practice questions
Question 1
Express as a fraction in simplest form.
$625^{\frac{-3}{4}}$625−34
Question 2
Express $\sqrt[5]{x^7}$5√x7 in index form.
Question 3
Fill in the blanks to simplify the given expression.
$\sqrt{m^8}$√m8 $=$= $\left(m^8\right)^{\editable{}}$(m8) $=$= $m^{\editable{}\times\frac{1}{2}}$m×12 $=$= $m^{\editable{}}$m
Question 4
Simplify $\left(625u^{16}v^{12}\right)^{\frac{1}{2}}$(625u16v12)12.
Question 5
Assuming that $y$y represents a non-negative number, simplify the following expression:
$\sqrt{16y^2+24y+9}$√16y2+24y+9