# Bah! I am lost with exponents!

• Jul 9th 2009, 05:02 PM
A Beautiful Mind
Bah! I am lost with exponents!
$\displaystyle \sqrt[4]{16a^7b^8}$$\displaystyle =$$\displaystyle 2ab^2 \sqrt[4]{a^3}$

I don't know how to make the jump from the first thing to the last thing.

EDIT: There was a typo, sorry.
• Jul 9th 2009, 05:25 PM
Kasper
$\displaystyle \sqrt[4]{16a^7b^8}$

$\displaystyle =\sqrt[4]{16a^3a^4b^8}$

$\displaystyle =\sqrt[4]{16}\sqrt[4]{a^3}\sqrt[4]{a^4}\sqrt[4]{b^8}$

$\displaystyle =2ab^2\sqrt[4]{a^3}$

Does this help?
• Jul 9th 2009, 05:53 PM
allyourbass2212
Quote:

Originally Posted by Kasper
$\displaystyle \sqrt[4]{16a^7b^8}$

$\displaystyle =\sqrt[4]{16a^3a^4b^8}$

$\displaystyle =\sqrt[4]{16}\sqrt[4]{a^3}\sqrt[4]{a^4}\sqrt[4]{b^8}$

$\displaystyle =2ab^2\sqrt[4]{a^3}$

Does this help?

I dont mean to intrude, but would someone mind expanding on this solution? Im not sure I quite understand the fine details of what is going on here.

Thanks
• Jul 9th 2009, 06:05 PM
Kasper
Quote:

Originally Posted by allyourbass2212
I dont mean to intrude, but would someone mind expanding on this solution? Im not sure I quite understand the fine details of what is going on here.

Thanks

For sure!

Well http://www.mathhelpforum.com/math-he...c2eb0c01-1.gif http://www.mathhelpforum.com/math-he...c84614d8-1.gif because $\displaystyle a^3a^4=a^7$ Just a small expansion.

$\displaystyle \sqrt[4]{16a^3a^4b^8}$http://www.mathhelpforum.com/math-he...32bb6d07-1.gif

By the rule $\displaystyle \sqrt{ab}=\sqrt{a}\sqrt{b}$, so we can split up the expression and take the fourth root of each term.

For additional clarification, I will rewrite that.

$\displaystyle \sqrt[4]{16}\sqrt[4]{a^3}\sqrt[4]{a^4}\sqrt[4]{b^8} $$\displaystyle =\sqrt[4]{2^4}\sqrt[4]{a^3}\sqrt[4]{a^4}\sqrt[4]{b^8}$$\displaystyle =2ab^2\sqrt[4]{a^3}$

Does this help, or should I clarify more?