How do I figure out the value of a number raised to a fractional power?

Question

For example, if I had something like $9^{1\over 2}$, how should I determine how many times I should use the base (the number being raised to the power of the exponent) as a factor? I just need something specific and readable. Should I use a root symbol or something? I'm asking this because I'm thinking it might be the opposite of a normal power.

Answer 1

Your thinking is on the right track: this is "opposite" a normal power in a way.

We can say $(9^\frac 12)^2 = 9,$ so $9^\frac 12$ is a solution for $x$ in the equation $x^2 = 9.$ The solutions are $x = \pm \sqrt{9}.$ By convention, we choose the positive square root, so we write $9^\frac 12 = \sqrt{9}.$

Answer 2

To complement David K, if you are asking in general, the convention is to write it

$$ x^{p/q}=\left(x^p\right)^{1/q}=\sqrt[q]{x^p} $$

BTW, in $\LaTeX$ you write \sqrt[q]{ } for that funny root $\sqrt[q]{\quad}$

Answer 3

Remember that $x^a \times x^b = x^{a+b}$.

So $(x^\frac{1}{2})^2 = x^\frac{1}{2} \times x^\frac{1}{2}=x^{\frac{1}{2}+\frac{1}{2}}=x^1=x$. Then taking the square root of both sides, $x^\frac{1}{2}=\pm \sqrt{x}$, and as David K says we take the positive root by convention.

Or you can use $(x^m)^n=(x^n)^m=x^{m \times n}$, which then lets you generalise further - first you show that $x^\frac{1}{n}=\sqrt[n]{x}$, and then you get $x^\frac{m}{n}=\sqrt[n]{x^m}=(\sqrt[n]{x})^m$ and so forth.