$\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.

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- Jul 9th 2009, 05:02 PMA Beautiful MindBah! 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 PMKasper
$\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 PMallyourbass2212
- Jul 9th 2009, 06:05 PMKasper
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?