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?

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- Apr 22nd 2008, 04:00 PM #1

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- Apr 22nd 2008, 04:57 PM #2
Start by factoring 64 down to it's prime factorization,

So 64 can be written as $\displaystyle 2^6$

Keep this in mind.

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

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

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

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

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

Also what is (n^3m^9)^1/2?

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

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

Or we can also do it this way,

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

These are all equivalent statements.

- Apr 22nd 2008, 07:09 PM #3

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- Apr 22nd 2008, 07:24 PM #4
Ah...

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

We can further simplify this by,

$\displaystyle

\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 #5

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