Maths question.

Hi I found this odd question in an exam paper and have never come across it before could You help me understand how to do it please?

Evaluate 16^(-3/2). (/ = fraction.)

You need to be aware of the facts that $a^{1/2}=\sqrt{a}$, $a^{-1}=\frac{1}{a}$, and that $(\sqrt{a})^n = \sqrt{a^n}$

So simplify your problem by expressing it using fractions and roots.

You have to change the 16^(-3/2) to 1/16^(3/2). Mathematically, 2^(-1) is same as 1/2^1.
So we can also write 1/16^(3/2) as 1/16^(3)x1/2. Anything to number 1/2 is the same as the square root of that number. So, 1/({√16}^3). The root of 16 is 4. Then we get 1/(4^3). And finally we get 1/64 or 64^-1 ( we have explained it in the second sentence of this writeup.
I hope you have gotten it.

I do not really understand what I would be achieving by simplifying or
what that would do. Could you explain what I would be doing to evaluate it as I'm really confused by what it means by this in terms of maths?

When it says 'evaluate' you can think of it as expressing a particular value in its simplest form.

For example, $4^{-1/2}$ is the same as $\frac{1}{4^{1/2}}$ which is the same as $\frac{1}{\sqrt{4}}$, and the simplest way to express this is to say it is the same as $\frac{1}{2}$, hence we'd say that $4^{-1/2}$ evaluates to $\frac{1}{2}$ which is the simplest form.

This is what you need to do with your question.

Why is 4^(-1/2) the same as root 4

It's not. It's the same as 1/root(4)

Is that not equal to 1/2?

Yes. It's what I said when I was working it out.

You have to change the 16^(-3/2) to 1/16^(3/2). Mathematically, 2^(-1) is same as 1/2^1.
So we can also write 1/16^(3/2) as 1/16^(3)x1/2. Anything to number 1/2 is the same as the square root of that number. So, 1/({√16}^3). The root of 16 is 4. Then we get 1/(4^3). And finally we get 1/64 or 64^-1 ( we have explained it in the second sentence of this writeup.
I hope you have gotten it.