# Bah! I am lost with exponents!

• July 9th 2009, 06:02 PM
A Beautiful Mind
Bah! I am lost with exponents!
$\sqrt[4]{16a^7b^8}$ $=$ $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.
• July 9th 2009, 06:25 PM
Kasper
$
\sqrt[4]{16a^7b^8}
$

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

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

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

Does this help?
• July 9th 2009, 06:53 PM
allyourbass2212
Quote:

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

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

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

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

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
• July 9th 2009, 07:05 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 $a^3a^4=a^7$ Just a small expansion.

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

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

For additional clarification, I will rewrite that.

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

Does this help, or should I clarify more?