Algebra Question/ Roots and Rational Exponents

[prodgers2]

I need to figure out how to solve the following problem: (2p)^1/2 * (2p^3)^1/3 can someone help me?

[JohnHarris]

I need to figure out how to solve the following problem: (2p)^1/2 * (2p^3)^1/3 can someone help me?

First you will need a hierarchy rule, that is what gets done first. Basically the one we would need is that exponentiation gets done before multiplication. Call it rule H1.

You will also need three distributive rules. One, call it D1, says
D1: (a^x)^y = a^(x*y)
Rule two, call it D2, says
D2: (a*b)^x = (a^x) * (b^x)
Rule 3, call it D3, says
D3: (a^x) * (a^y) = a^(x+y)

Oh, I almost forgot the associative rule for multiplication, i.e. a * b = b * a, which I assume is true for the problems here.

I'll assume 1/2 is one-half and 1/3 is one third.
(2p)^1/2 means (2 * p)^1/2 = (2^1/2) * (p^1/2) by rule D2

By rule H1
(2p^3) = (2 * p^3)
so
(2p^3)^1/3 = 2^1/3 * (p^3)^1/3 by rule D2
= 2^1/3 * p^(3*1/3) = 2^1/3 * p^1 by rule D1

Note that p is the same as p^1 and use the associative rule several times to get
(2p)^1/2 * (2p^3)^1/3 = (2^1/2) * (p^1/2) * 2^1/3 * p
=2^1/2 * 2^1/3 * p^1/2 * p

Finally, use rule D3 to get
=2^1/2 * 2^1/3 * p^1/2 * p = 2^(1/2 + 1/3) * p^(1/2 + 1)
= 2^5/6 * p^3/2

Note: IF (2p^3)^1/3 was supposed to be ((2p)^3)^1/3 that would mean a different (and much simpler) answer but the mechanics of the solution would be the same. Hint: let a = 2p

[tim]

Thanks John for a very thorough response. FYI a*b = b*a is the commutative property, not the associative property. Don’t sweat it though; I’ve even seen some of our instructors make this mistake, and the term itself is not important to getting the question correct..