rewriting rational exponents

Question

A naive question that's been confusing me and that I didn't see answered elsewhere:

Can I always write a fraction $\frac{a}{b}$ as $\ a (b)^{-1}$?

And if so, what about the following:

$$\ x^{\frac{a}{b}} = x^{a(b)^{-1}} = x^{a(-b)} = x^{-ab} = \frac{1}{x^{ab}}$$

Plugging in e.g. $\ a=1, b=2, x=9$, I quickly see that this must be wrong as $\ 9^{\frac{1}{2}}=3\neq \frac{1}{81}=\frac{1}{9^{2}}$.

However, what exactly goes wrong?

Answer 1

Hint:

$b^{-1} \ne -b$

so your conclusion is wrong.

Answer 2

we have $$\frac{a}{b}=ab^{-1}$$ thus we have $$x^{\frac{a}{b}}=x^{ab^{-1}}$$ thats all

Answer 3

The parts marked red are not equal:

$$x^{\frac{a}{b}}=x^{\color\red{a(b)^{-1}}}=x^{\color\red{a(-b)}}=x^{-ab}=\frac{1}{x^{ab}}$$