# Simplify using Exponent Laws

• Sep 12th 2007, 11:31 AM
Jeavus
Simplify 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 AM
Jhevon
Quote:

Originally Posted by Jeavus
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

$\displaystyle \frac {3t - 2t^{-1}}{t^3} = \frac {3t}{t^3} - \frac {2t^{-1}}{t^3}$

$\displaystyle = 3 t^{1 - 3} - 2t^{-1-3}$

Now simplify that.

The second question is done similarly
• Sep 12th 2007, 11:37 AM
Jeavus
Can you explain how you got:

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

Thanks. :)

EDIT: Nevermind, sorry.
• Sep 12th 2007, 11:40 AM
Jhevon
Quote:

Originally Posted by Jeavus
Can you explain how you got:

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

Thanks. :)

There is a law of exponents that says $\displaystyle \frac {x^a}{x^b} = x^{a - b}$

so, for instance, $\displaystyle \frac {3t}{t^3} = 3 \cdot \frac {t}{t^3} = 3 t^{1 - 3}$, since $\displaystyle t = t^1$
• Sep 12th 2007, 11:47 AM
Jeavus

(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 AM
Jhevon
Quote:

Originally Posted by Jeavus

(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?

those don't really look simpler to me. you eliminated some of the variables, but the ones that remain have even higher powers than before. i would leave them separated as sums
• Sep 12th 2007, 12:00 PM
Jeavus
How do I go about solving a problem like this?

1) (4 - (square root)x)/x^(3/2)
• Sep 12th 2007, 12:11 PM
Jhevon
Quote:

Originally Posted by Jeavus
How do I go about solving a problem like this?

1) (4 - (square root)x)/x^(3/2)

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 PM
Jeavus
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 PM
Jhevon
Quote:

Originally Posted by Jeavus
And do you know how to go abouts solving these two? Last two I promise. :)

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

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

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 PM
Jeavus
Thanks so much for all your help. :)

PS: I did what you requested! :P
• Sep 12th 2007, 12:41 PM
Jhevon
Quote:

Originally Posted by Jeavus
Thanks so much for all your help. :)

PS: I did what you requested! :P

ok, just note that we have $\displaystyle x^{{\color {red} -} 1/2} = \frac {1}{x^{1/2}}$ in the denominator, not $\displaystyle x^{1/2}$, so you can't just cancel. more simplification is needed before you cancel anything