Arithmetic : Powers and Roots Question

**dumluck** · January 6th 2011, 04:09 AM

Hi All,
I'm new to the forum so firstly hello. The following question may be rudimentary given the level of ability of some of you but I would very much appreciate a clear understanding/answer.

1. Can someone help me understand how 2√27 simplifies to (2 X 3 X √3).
2. Can someone help me understand how 3^√27 X 3^√3 simplifies to 3^3√3

I've read quite a few of the web manuals on Powers and Roots but these specific examples are a block for me. If anyone can point me towards a comprensive resource I'd very much appreciate it.

Thanks in Advance.

**Plato** · January 6th 2011, 04:25 AM

Do you understand that $\sqrt{27}=\sqrt{3^2\cdot 3}=\sqrt{3^2}\cdot\sqrt{3}~?$

**Prove It** · January 6th 2011, 04:27 AM

$\displaystyle 3^{\sqrt{27}} \times 3^{\sqrt{3}} = 3^{\sqrt{27} + \sqrt{3}}$ .

Can you simplify $\displaystyle \sqrt{27}$ ?

**dumluck** · January 6th 2011, 04:34 AM

Originally Posted by Plato

Do you understand that $\sqrt{27}=\sqrt{3^2\cdot 3}=\sqrt{3^2}\cdot\sqrt{3}~?$

Hi Plato:
Thanks for your reponse. I get that $\sqrt{3^2\cdot 3}=\sqrt{3^2}\cdot\sqrt{3}~?$ , I don't know how one derives the first bit: $\sqrt{27}=\sqrt{3^2\cdot 3}$

**Prove It** · January 6th 2011, 04:38 AM

Because $\displaystyle 9\cdot 3 = 27$ ... You need to look for square factors. In this case, you can see that $\displaystyle 3^2 = 9$ is a factor.

**dumluck** · January 6th 2011, 04:49 AM

Originally Posted by Prove It

Because $\displaystyle 9\cdot 3 = 27$ ... You need to look for square factors. In this case, you can see that $\displaystyle 3^2 = 9$ is a factor.

Removed. figured it out. So just to post the solution...

1. Can someone help me understand how 2√27 simplifies to (2 X 3 X √3).
1a. Factor of 27 is 3 * 9. 9 is 3^2
1b. Simplifies to 2√3^2 X 2 = 2√3^2 * √3
1c. √3^2 = 3
1d. Leaving 2 * 3 * √3

2. Can someone help me understand how 3^√27 X 3^√3 simplifies to 3^3√3
2a. Factor of 27 is 3 * 9. 9 is 3^2
2b. Simplifies to 3 X √3 * 9
2c. Simplifies to 3 X √3 * √3^2
2d. Leaving 3 X 3 * √3

Thanks to all.

**DrSteve** · January 6th 2011, 05:11 AM

I would review how to get the prime factorization of a number using factor trees. For example, do you see why

$3000 = 2^3\cdot 3 \cdot 5^3?$

Now, you can take the square root of 3000 as follows:

$\sqrt{3000} =\sqrt{2^3\cdot 3 \cdot 5^3}=\sqrt{2^2\cdot 5^2}\cdot \sqrt{2\cdot 3 \cdot 5}=2\cdot 5\sqrt{30}=10\sqrt{30}$

This is a systematic procedure that will always get you the answer. After you're comfortable with this you can begin to use shortcuts such as

$\sqrt{3000} =\sqrt{100\cdot 30}=10\sqrt{30}$

Note that I know that $\sqrt{30}$ can't be reduced further because the prime factorization of 30 has no exponents greater than 1.

Perhaps someone who is better at latex than me can draw a factor tree for 3000?

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