I have two questions, if any help is given I'd appreciate it.

Simplify using the laws of exponents.

1) (3t - 2t^(-1)) / t^3

2) (3p^2 - p^(-3)) / p^4

Thank you very much. :)

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- Sep 12th 2007, 11:31 AMJeavusSimplify using Exponent Laws
I have two questions, if any help is given I'd appreciate it.

Simplify using the laws of exponents.

1) (3t - 2t^(-1)) / t^3

2) (3p^2 - p^(-3)) / p^4

Thank you very much. :) - Sep 12th 2007, 11:33 AMJhevon
- Sep 12th 2007, 11:37 AMJeavus
Can you explain how you got:

3 t^{1 - 3} - 2t^{-1-3}

Thanks. :)

EDIT: Nevermind, sorry.

- Sep 12th 2007, 11:40 AMJhevon
- Sep 12th 2007, 11:47 AMJeavus
So my final answers were:

(3t - 2t^(-1)) / t^3

= 3t / t^3 - 2t^(-1)/t^3

= 3t^(1-3) - 2t^(-1-3)

= 3t^(-2) - 2t^(-4)

= (3/t^2) - (2/t^4)

= (3t^2 - 2)/t^4

Is that correct?

(3p^2 - p^(-3))/p^4

= 3p^(2-4) - p^(-3-4)

= 3p^(-2) - p^(-7)

= (3/p^2) - (1/p^7)

= (3p^5 - 1) / p^7

Is that correct? - Sep 12th 2007, 11:52 AMJhevon
- Sep 12th 2007, 12:00 PMJeavus
How do I go about solving a problem like this?

1) (4 - (square root)x)/x^(3/2) - Sep 12th 2007, 12:11 PMJhevon
you do this the same as the last two, except you must know that $\displaystyle \sqrt {x} = x^{\frac {1}{2}}$

see post #3 here to learn how to change various root-forms into powers. - Sep 12th 2007, 12:25 PMJeavus
And do you know how to go abouts solving these two? Last two I promise. :)

1) (x-9) / (x^(1/2)-3)

2) (x-1)/((square root)x - x) - Sep 12th 2007, 12:31 PMJhevon
Notice that the top is the difference of two squares in terms of $\displaystyle \sqrt {x}$

$\displaystyle \frac {x - 9}{x^{1/2} - 3} = \frac {\left( \sqrt {x} \right)^2 - 3^2}{\sqrt {x} - 3}$

$\displaystyle = \frac { \left( \sqrt {x} + 3 \right) \left( \sqrt {x} - 3 \right)}{\sqrt {x} - 3}$

Hopefully the last line is obvious to you

Quote:

2) (x-1)/((square root)x - x)

how do you think you should proceed?

Hint: You do the same trick with the top as I did in the last question - Sep 12th 2007, 12:36 PMJeavus
Thanks so much for all your help. :)

PS: I did what you requested! :P - Sep 12th 2007, 12:41 PMJhevon