# rational exponents? Solve please

• Jun 2nd 2012, 06:36 AM
zbest1966
rational exponents? Solve please
(-3a1/4)(9a)-3/2
• Jun 2nd 2012, 06:53 AM
Reckoner
Re: rational exponents? Solve please
Quote:

Originally Posted by zbest1966
(-3a1/4)(9a)-3/2

Here are some properties you should know:

$a^ma^n = a^{m+n}$

$(ab)^n = a^nb^n$

$a^{m/n} = \sqrt[n]{a^m}$
• Jun 2nd 2012, 07:03 AM
zbest1966
Re: rational exponents? Solve please
I understand the properties but how do you solve it.
• Jun 2nd 2012, 07:25 AM
Reckoner
Re: rational exponents? Solve please
Quote:

Originally Posted by zbest1966
I understand the properties but how do you solve it.

If you understood the properties, I suspect you would not be having difficulties with the problem. By the way, we have nothing to "solve" here, this is not an equation. Rather, we are "simplifying" the expression.

$\left(-3a^{1/4}\right)(9a)^{-3/2}$

$=-3a^{1/4}\cdot9^{-3/2}a^{-3/2}$

$=-\frac{3a^{1/4}}{9^{3/2}a^{3/2}}$

$=-\frac{3a^{1/4}}{27a^{3/2}}$

Now, use the properties $\frac{a^m}{a^n} = a^{m-n}$ and $a^{-n} = \frac1{a^n}$ to finish it:

$=-\frac{a^{-5/4}}{9}=-\frac1{9a^{5/4}}$

As a final step, you may want to convert $a^{5/4}$ to radical form, depending on the problem's instructions.
• Jun 2nd 2012, 07:51 AM
HallsofIvy
Re: rational exponents? Solve please
Quote:

Originally Posted by zbest1966
I understand the properties but how do you solve it.

You could actually apply the properties you say you understand or you could ask someone else to solve it for you. I see you chose the second solution method.