algerba 2 help today?

Note that x^(4/5) = x * x^(-1/5) and x^(3/4) = x * x^(-1/4) so

16x^(4/5) + 8x^(3/4) = 2x*[8x^(-1/5) + 4x^(-1/4)]

16x^(4/5) + 8x^(3/4)
= 2[8x^(4/5)] + 2[4x^(3/4)]
= 2x[8x^(4/5 - 1)] + 2x[4x^(3/4 - 1)]
= 2x[8x^(-1/5)] + 2x[4x^(-1/4)]
= 2x[8x^(1/5) + 4x^(-1/4)]

http://www.flickr.com/photos/dwread/3632302485/

when multiplying say x*x you add the exponent so you get x^2
same goes for exponents that are fractions x*x^-1/5 = x^4/5

so for you problems pull out a 2x

2x^(1)[ 8x^(-1/5) + 4x^(-1/4)] i added the ^1 to help visually
so

2x^1 * 8x^-1/5 =(2*8)x^(1 + -1/5) = 16x^4/5

2x^1 * 4x^-1/4 = (2*4)x^(1 - 1/4) = 8x^(3/4)

hope this helps

Well, 16 = 2 * 8 and 8 = 2 * 4

That's the easy part.

For the next part, understand that x = x^1 and when you multiply two numbers together you add their exponents.

x^2 * x^3 = x^(2 + 3) = x^5

x^(4/5) can be thought of as x^(-1/5 + 1)
x^(3/4) can be thought of as x^(-1/4 + 1)

So 16x^(4/5) / 2x = 8x^(-1/5)
And 8x^(3/4) / 2x = 4x^(-1/4)

So 2x * (8x^(-1/5) + 4x^(-1/4))

Why you would want to do that is beyond me. I'd pull out a common factor of 8 and leave the rest. Fraction exponents are sucky enough without complicating them in this fashion.

16x^5 factor out 2x and we have 2x (8x^(1/5))

8x^(3/4) factor out and we have 2x (4x^(-1/4))

x = x^1, so when you factor out x^1, you just subtract the exponents, like for x^5, it's x^(5 - 1) = x^4

for x^(3/4), it's x^(3/4 - 1) = x^(-1/4)

So, you have 16x^(4/5) + 8x^(3/4) = 2x (8x^(1/5) + 4x^(-1/4))