Could anyone help me with these equations?

[Hanajima]

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:

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(I'm still working on 4 pages)

 

[galactus]

I will show you how to do g. It has several exponent laws to use.

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

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.

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

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

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

\(\displaystyle =\frac{36x^{6}y^{16}}{4x^{10}y^{10}}=\frac{9y^{6}}{x^{4}}\)

 

[whitecanvas]

hi there,

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

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

Abby

 

[Deleted member 4993]

hi there,

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

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

Abby

Anotherway:

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

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

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

\(\displaystyle |x| \, = 2^9\)

 

[galactus]

For 13b,

\(\displaystyle \sqrt[5]{-32^{6}}=\sqrt[5]{-(2^{5})^{6}}=\sqrt[5]{-2^{30}}=-2^{6}=-64\)


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. \(\displaystyle \frac{5a}{\sqrt[5]{8a^{9}b^{11}}}\)

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

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

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

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

\(\displaystyle \frac{5a\sqrt[5]{4ab^{4}}}{2a^{2}b^{3}}=\frac{5\sqrt[5]{4ab^{4}}}{2ab^{3}}\)

 

[mmm4444bot]

… 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