Fractional Exponents - Is the sign discarded?

Question

For example,

16^(3/4)

Is the accepted as both -8 and 8 or just 8?

I ask this because on an AS maths mark scheme it says to condone -8

Thanks

Answer 1

There is a common missunderstanding of the function square-root, that often produce confusions when talking about fractional powers with an even denominator. First of all, note that the domain of $f(x)=\sqrt{x}$ is the non-negative real numbers, and the possible values it takes are always positive. So, for instance, $\sqrt{4}=2$, or $\sqrt{9}=3$.

However, a different thing is to ask about the possible solutions of the equation $x^2=4$, which are two different values: $x=2=\sqrt{4}$ and $x=-\sqrt{4}=-2$. Note that in the last equation the minus sign is outside of the square root, which indicates that the value of $\sqrt{4}$ is positive.

Usually, when solving the equation $x^2=9$ we say $x=\pm \sqrt{9}=\pm 3$, but this does not mean that the square root of $9$ is both positive and negative, but rather that there are two solutions of the equation, on if $\sqrt{9}$ and the other one is $-\sqrt{9}$ (a completely different number).

So, it is not that both $8$ and $-8$ are equal to $16^{3/4}$ (which would imply that $-8=8$, and this is absurd!). In this case we have:

$16^{3/4}=(16^{1/2})^{3/2}=(\sqrt{16})^{3/2}=4^{3/2}=(\sqrt{4})^3=2^3=8$.

Answer 2

After reading Wore's answer, I feel it is worth it to represent another direction of the argument.

First of all, this step simplifies our problem: $$16^{3/4}=(16^3)^{1/4}=4096^{1/4}$$

Now, the problem is how we define this.

Generally, we define $a^{1/n}=\sqrt[n]a$, but we don't always define it this way.

Sometimes, we define $a^{1/n}=x$, or $a=x^n$. Then we algebraically solve for $x$ with polynomials and stuff, yielding many complex solutions.

Other times, $a^{1/n}$ is defined as $+\sqrt[n]a$, yielding one real solution.

It really depends on the context of how we are to go about solving such a problem, which I imagine, for your case, would be to use $+\sqrt[n]a$.

Sometimes, however, we use the 'algebraic' path, which in this case, yields $x=8,-8,8i,-8i$.

$$(x-8)(x+8)(x-8i)(x+8i)=x^4-4096=0$$

We usually say $x^n=a\implies x=a^{1/n}$, not $x=\sqrt[n]a$. If the context uses the exponent form, it is possible that multiple answers are to expect. If it is in radical form, one answer is to expect.