Could anyone help me with these equations?

Hanajima

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Aug 27, 2009
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I wasn't exactly too sure where to put my question, but this was a page from my summer Calculus Packet, so I decided to post it here.
I really don't understand how to do these problems or where to begin. I'm not even sure I've even learned them before =x.
I was hoping someone could explain how to go about solving these problems for me. Any help that you guys could offer me would be great, thank you!

Work Proof:

Page 1: http://i71.photobucket.com/albums/i140/hanae_sama/Work Proof/WorkProof008.jpg

Page 3: http://i71.photobucket.com/albums/i140/hanae_sama/Work Proof/WorkProof009.jpg

http://i71.photobucket.com/albums/i140/hanae_sama/Work Proof/WorkProof010.jpg

Page 4: http://i71.photobucket.com/albums/i140/hanae_sama/Work Proof/WorkProof011.jpg

Page 7: http://i71.photobucket.com/albums/i140/hanae_sama/Work Proof/WorkProof012.jpg

Page 9: http://i71.photobucket.com/albums/i140/hanae_sama/Work Proof/WorkProof013.jpg

(I'm still working on 4 pages)
 
I will show you how to do g. It has several exponent laws to use.

Remember that xa=1xa\displaystyle x^{-a}=\frac{1}{x^{a}}

So, whenever you see a negative exponent, put it on the other side. If it's in the denominator, put it in the numerator and vice versa.

(2x3y56x5y8)2\displaystyle \left(\frac{2x^{-3}y^{5}}{6x^{-5}y^{8}}\right)^{-2}

Remember the law: (am)n=amn\displaystyle (a^{m})^{n}=a^{mn} and aman=amn\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n} and aman=am+n\displaystyle a^{m}\cdot a^{n}=a^{m+n}

(2x3y56x5y8)2=22x6y1062x10y16\displaystyle \left(\frac{2x^{-3}y^{5}}{6x^{-5}y^{8}}\right)^{-2}=\frac{2^{-2}x^{6}y^{-10}}{6^{-2}x^{10}y^{-16}}

=36x6y164x10y10=9y6x4\displaystyle =\frac{36x^{6}y^{16}}{4x^{10}y^{10}}=\frac{9y^{6}}{x^{4}}
 
hi there,

hope this helps with some of the problems you're having trouble with. good luck. i'm currently working on these too. :D

[attachment=0:2p7frv27]13.png[/attachment:2p7frv27]

Abby
 

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whitecanvas said:
hi there,

hope this helps with some of the problems you're having trouble with. good luck. i'm currently working on these too. :D

[attachment=0:3uuzguhy]13.png[/attachment:3uuzguhy]

Abby
Anotherway:

x23=64\displaystyle x^{\frac{2}{3}} \, = 64

x23=26\displaystyle x^{\frac{2}{3}} \, = 2^6

(x23)32=(26)32\displaystyle (x^{\frac{2}{3}})^{\frac{3}{2}} \, = (2^6)^{\frac{3}{2}}

x=29\displaystyle |x| \, = 2^9
 
For 13b,

3265=(25)65=2305=26=64\displaystyle \sqrt[5]{-32^{6}}=\sqrt[5]{-(2^{5})^{6}}=\sqrt[5]{-2^{30}}=-2^{6}=-64


Here is an example of rationalizing that gets rid of a radical in the denominator. The idea is to multiply by a power that gives a perfect power so the radical is eliminated.

i.e. 5a8a9b115\displaystyle \frac{5a}{\sqrt[5]{8a^{9}b^{11}}}

Multiply top and bottom by 4ab45\displaystyle \sqrt[5]{4ab^{4}}. When we do this it gets rid of the radical in the denominator.

5a8a9b1154ab454ab45\displaystyle \frac{5a}{\sqrt[5]{8a^{9}b^{11}}}\cdot \frac{\sqrt[5]{4ab^{4}}}{\sqrt[5]{4ab^{4}}}

We get 5a4ab4532a10b155\displaystyle \frac{5a\sqrt[5]{4ab^{4}}}{\sqrt[5]{32a^{10}b^{15}}}

See in the denominator everything is a perfect 5th root:

5a4ab452a2b3=54ab452ab3\displaystyle \frac{5a\sqrt[5]{4ab^{4}}}{2a^{2}b^{3}}=\frac{5\sqrt[5]{4ab^{4}}}{2ab^{3}}
 
whitecanvas said:
… root both sides to get x by itself …


Not quite, Abby.

The square root of x^2 is |x|.

The solutions are |x| = 512.

If you would like to report solutions without using absolute-value symbols (i.e., "get x by itself"), then you could write one of the following.

x = -512 or x = 512

OR

x = ± 512

OR

x ? {-512, 512}

Cheers 8-)

 
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