Can someone please explain the steps to get from -((9a)^1/2) to 3^-1a^-1/2

Question

anonymous · Answer

$$
-(9a)^{1\over2}=-\sqrt{9a}=-\sqrt{9}\times\sqrt{a}=-\sqrt{3^2}\times a^{1\over2}
=-3 a^{1\over2}
$$

anonymous · Answer

did you write the question correctly?

anonymous · Answer

Ok, and that can be taken one step further to $$3^{-1}a ^{-_{2}^{1}}$$

anonymous · Answer

I'm pretty sure I did. I'm simplifying an exercise in my text book, and that is the answer, I was just unsure how they got to it.  + and - mixed with exponents and radicals still confuse me

anonymous · Answer

nope.. remember$$a^{-m}={1\over a^m}$$
so, the given way, it cannot happen. (unless more information about "a" or something is provided)

anonymous · Answer

by the above rule, we can write it as:
$$-3a^{1\over2}=\frac{-1}{3^{-1}a^{-1\over2}}$$

anonymous · Answer

ok, what about if I forgot to tell you that this part of the question was the denominator and the initial   - is because it was moved up to the numerator?

anonymous · Answer

that sort of fits with the rule you just showed me, doesn't it?

anonymous · Answer

SEE!!!
math dont lie

anonymous · Answer

yep.

anonymous · Answer

sheeesh, I don't think I'll ever get this sorted in my head...

anonymous · Answer

Thank you for your help :D

anonymous · Answer

yer welcome <:)