I just started my first college calc class, and I cannot remember anything from Algebra....

What I need to know right now is.....

So I have (81x^-4 * y^8)^-1/4 * (xy^1/4)
All I need to do is simplify.
This is what I went to.

((x^4/81y^8))^1/4 * (xy^1/4)

I'm not sure if this is right so far. If it is, I don't know what to do next. I was thinking that I could distribute the fourth root to each thing inside the parentheses (if I can do that??) so it would be (x/3y^2) * (xy^1/4)

Anyway, that's all I really need to know right now.
Your help is appreciated, thanks!

fyi, if you're wondering whether two expressions are equal, i recommend using a calculator and picking some values for x and y (eg: 2 and 3), sticking it into both expressions, and seeing if it's equal. there's a chance for false positives (eg your numbers worked out coincidentally) but if the two expressions come out with different values then they're not equivalent.

We aren't supposed to use calculators, and I don't have one anyway. I guess I could use wolfram alpha, but I would still like to know if I did it correctly.

(81x^-4 * y^8)^-1/4 * (xy^1/4)

(ab)^x = a^x * b^x

Then: (81x^-4)^-1/4 * (y^8)^-1/4

a^x^y = a^(xy)

Then: 81^-1/4 * x^(-4*-1/4) = 81^-1/4 * x and y^(8*-1/4) = y^-2

81^-1/4 * x * y^-2 * x * y^1/4 = 81^-1/4 * x^2 * y^(-2+1/4)

Answer: 81^-1/4 * x^2 * y^(-1.75)

Uhm.... Where did you get the 1/4 from an exponent to multiplication?

(x/3y^2) * (xy^1/4)

Okay, So this is correct, but it needs to be in exponential notation. How would I do that?

(x^2)/(3y^(7/4))

That's it? I thought I tried that... I'll try again.