[SOLVED] Simplifying/Solving Rational Exponents

**ria3** · March 21st 2008, 04:20 PM

I'm having trouble simplifying this:
(2x^-2)^2(3x^4)^-3
the answer is 4/27x^16
but I have no idea how they got this??
Also, solving for 3a^(4/5)=48 could you please guide me through the steps?
thank you so much!

**ria3** · March 21st 2008, 04:32 PM

sorry, forgot to show my steps so far:
for the first qu:
I have it up to: 1/(4x^4)(-27x^12)
I don't understand how they got a 4 over 27x^16? I would think it would be a 1 over 27x^16.

For the second question:
I have 5sqrt(3a^4) = 48
and then I'm sorta stuck... could it be: 5*4=20, so (a^20)/3 = 16?

**teuthid** · March 21st 2008, 04:48 PM

use the exponent laws $\left(x^{a}\right)^{b}=x^{ab}$ and $(xy)^{a}=x^{a}y^{b}$ :

$\left(2x^{-2}\right)^{2}\left(3x^4\right)^{-3}=2^2x^{-4}3^{-3}x^{-12}$

then remember that $x^{a}x^{b}=x^{a+b}$ :

$=2^2\cdot 3^{-3}x^{-16}$

now recall that $x^{-a}=\left(\frac{1}{x^{a}}\right)$ . Also note that $3^{-3}$ is a positive number:

$=2^2\left(\frac{1}{3^3}\right)\left(\frac{1}{x^{16 }}\right)=\frac{4}{27x^{16}}$

**ria3** · March 21st 2008, 05:12 PM

thank you tuephid!
and nvm 'bout the 2nd one... i got it

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