# Thread: [SOLVED] division of an algebric expression

1. ## [SOLVED] division of an algebric expression

Please check if this is right:

$\frac {2^{-3}a^ \frac {2}{3} b^4}{2^{-2}a^ \frac {1}{6} b^4}$

$1^{-3-(-2)} a^{\frac {2}{3}-\frac {1}{6}} b^{4-4}$

$1^{-1} a^{\frac {4}{6}-\frac {1}{6}}$

$1 a^{\frac {3}{6}}$

$a^{\frac {3}{6}}$

$a^{\frac {1}{2}}$

$\sqrt{a}$

The exposants can't be fractions for the answer to be right...

(wow this was tough code to write!!!)

I'm really starting to hate math! sorry

2. Hello, Neenoon!

One correction . . .

$\frac {2^{\text{-}3}\,a^{\frac{2}{3}}\,b^4}{2^{-2}\,a^{\frac{1}{6}}\,b^4} \;\;=\;\;{\color{red}2}^{\text{-}3-(\text{-}2)}\,a^{\frac{2}{3}-\frac {1}{6}}\,b^{4-4} \;\;=\;\;2^{-1} a^{\frac {4}{6}-\frac {1}{6}}b^0 \;\;=\;\;\frac{1}{2}\,a^{\frac{1}{2}} \;\;=\;\;\frac{1}{2}\sqrt{a}$

3. Thank you. I don't understand though, how come the 2 devided by 2, still stays 2?

4. Originally Posted by Neenoon
Thank you. I don't understand though, how come the 2 devided by 2, still stays 2?
It's NOT 2 divided by 2. Don't ignore the exponent in this case. Here's the rule:

$\frac{x^a}{x^b}=x^{a-b}$

You see. The x (base) stayed the same. We only subtracted the exponents.

5. Originally Posted by masters
It's NOT 2 divided by 2. Don't ignore the exponent in this case. Here's the rule:

$\frac{x^a}{x^b}=x^{a-b}$

You see. The x (base) stayed the same. We only subtracted the exponents.
Really!?!? Hummmm.... I better check all my homework for such a case!