# Rational Exponents

• Sep 16th 2007, 02:38 PM
Ash
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.
• Sep 16th 2007, 02:43 PM
TKHunny
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.
• Sep 16th 2007, 03:04 PM
Ash
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?
• Sep 16th 2007, 03:19 PM
TKHunny
You have the rules, don't you?

\$\displaystyle a^{b}*a^{c} = a^{b+c}\$

That's all it takes.

2/5 + 8/5 = 10/5 = 2
• Sep 16th 2007, 03:48 PM
Ash
Quote:

Originally Posted by TKHunny
You have the rules, don't you?

\$\displaystyle a^{b}*a^{c} = a^{b+c}\$

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