Rational Exponents

I need help in learning how to do these two problems.

#1 (3^1/2 2^3/2)^2

#2 b^2/5 (b^8/5 + b 3/5 ) *Assuming that b is positive

I know the answer to #1 is 24 but, the only way I obtained a 24 was by:

(3^1/2 2^3/2)^2
3^1 2^3
(3) 8 = 24, but, I thought, you're suppose to square the 3 & 2 and change it into (9^1 & 4^3) I'm confused by both problems.

Can you show me both steps so I can see where I am going wrong and how to obtain the answers.

Quote:

---

Originally Posted by Ash

I thought, you're suppose to square the 3 & 2 and change it into (9^1 & 4^3)

---

Why would you use the exponent twice? You DID use the outside '2' when you modified the internal exponents.

[3^(1/2)]^2 = 3^1

[2^(3/2)]^2 = 2^3

There is no more 2 in the exponent.

Quote:

---

Originally Posted by TKHunny

Why would you use the exponent twice? You DID use the outside '2' when you modified the internal exponents.

[3^(1/2)]^2 = 3^1

[2^(3/2)]^2 = 2^3

There is no more 2 in the exponent.

---

Thanks, that makes sense, Do you know the steps to solving the second one?

You have the rules, don't you?



That's all it takes.

2/5 + 8/5 = 10/5 = 2

Quote:

---

Originally Posted by TKHunny

You have the rules, don't you?



That's all it takes.

2/5 + 8/5 = 10/5 = 2

---

So you first add the 2/5 + 8/5 = 2, then add the 2/5 + 3/5 =1

So, the answer is, b^2 + b



Thanks again for your help.

yeah you just need to study it , its really easy