# Please Help

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• Apr 22nd 2008, 04:00 PM
anime_maniac20
Please Help
I have to learn how do roots without a calculator and one of the practice problems i get was 12 root of 64 equal square root of two.How could i figure out problems like these without a calculator?Also what is (n^3m^9)^1/2?
• Apr 22nd 2008, 04:57 PM
Jen
Quote:

Originally Posted by anime_maniac20
I have to learn how do roots without a calculator and one of the practice problems i get was 12 root of 64 equal square root of two.

Start by factoring 64 down to it's prime factorization,
Attachment 5991

So 64 can be written as $2^6$

Keep this in mind.

So by the definition of roots, $\sqrt[12]{64}=64^{\frac{1}{12}}$

Now substituting the prime factorization of 64 in place of 64, and writing it as an exponent as previously shown we get,

$\left(2^6\right)^{\frac{1}{12}}$

Using the law for the power of a power, we multiply the exponents together which gives us,

$2^{\frac{6}{12}}=2^{\frac{1}{2}}=\sqrt{2}$

Quote:

Also what is (n^3m^9)^1/2?
$\left(n^3m^9\right)^{\frac{1}{2}}$

Using the same law of exponents the outside exponent gets "distributed" to each inside term.

$n^{3*\frac{1}{2}}m^{9*\frac{1}{2}}=n^{\frac{3}{2}} m^{\frac{9}{2}}=\sqrt{n^3}\sqrt{m^9}$

Or we can also do it this way,

$\sqrt{n^3m^9}=\sqrt{n^3}\sqrt{m^9}$

These are all equivalent statements.
• Apr 22nd 2008, 07:09 PM
anime_maniac20
Quote:

Using the same law of exponents the outside exponent gets "distributed" to each inside term.

Or we can also do it this way,

I had the exact answers but the answer that was correct was like $nm^{4}\sqrt{nm}$
• Apr 22nd 2008, 07:24 PM
Jen
Quote:

Originally Posted by anime_maniac20
I had the exact answers but the answer that was correct was like $nm^{4}\sqrt{nm}$

Ah...

That is correct, there was one step of simplification that I missed, thank you.

We can further simplify this by,

$
\sqrt{n^3m^9}=\sqrt{n^3}\sqrt{m^9}=\sqrt{n^2*n}\sq rt{m^8*m}=nm^4\sqrt{nm}
$

I skipped a few steps there where I used the rules I previously discussed. Let me know if you follow this o.k.
• Apr 22nd 2008, 08:08 PM
anime_maniac20
Thank you so much,I did understand it. You mad it look really simple.(Happy)