If k, n, x and y are positive numbers satisfying $\displaystyle x^\frac{-4}{3}=k^{-2}$ and $\displaystyle y^\frac{4}{3}=n^2$, what is $\displaystyle (xy)^\frac{-2}{3}$ in terms of n and k?

Thanks in advance!! (:

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- Sep 28th 2008, 04:31 AMfabxx[SOLVED] SAT math page 674 #19
If k, n, x and y are positive numbers satisfying $\displaystyle x^\frac{-4}{3}=k^{-2}$ and $\displaystyle y^\frac{4}{3}=n^2$, what is $\displaystyle (xy)^\frac{-2}{3}$ in terms of n and k?

Thanks in advance!! (: - Sep 28th 2008, 04:42 AMskeeter
this will get you started ...

$\displaystyle x^{-\frac{4}{3}} = k^{-2}$

$\displaystyle \left(x^{-\frac{4}{3}}\right)^{-\frac{3}{4}} = \left(k^{-2}\right)^{-\frac{3}{4}}$

$\displaystyle x = k^{\frac{3}{2}}$

now do the same to solve for y in terms of n ... then evaluate $\displaystyle (xy)^{-\frac{2}{3}}$ - Sep 28th 2008, 05:42 AMfabxx
- Sep 28th 2008, 05:52 AMMoo
- Sep 28th 2008, 06:05 AMskeeter
my apologies ... I assumed you had some knowledge regarding the Laws of Exponents

- Sep 28th 2008, 06:42 AMshailen.sobhee
Basically,

If its SATs,, forget about complex rules.*AS A RULE*

$\displaystyle x ^\frac{-4}{3} = k^{-2} $ ...(1)

and

$\displaystyle y^\frac{4}{3} = n^2$ ... (2)

From (1),

$\displaystyle \frac{1}{x ^\frac{4}{3}} = \frac{1}{k^2}$

Cross multiply and shift the x-term to the left hand side, you get:

$\displaystyle x ^\frac{4}{3} = k^2$ ... (3)

Multiply (2) and (3)

$\displaystyle x ^\frac{4}{3} * y^\frac{4}{3} = n^2* k^2$

$\displaystyle (xy) ^\frac{4}{3} = (nk)^2$

$\displaystyle ((xy)^\frac{2}{3})^2 = (nk)^2$

Squaring on both sides and making a reciprocal of the left-hand and the right-hand-side term gives:

$\displaystyle (xy)^\frac{-2}{3}=\frac{1}{nk}$ <= Answer - Sep 28th 2008, 06:46 AMfabxx
i got x=k^(8/3) and y=n^(3/2)

If i substitute I have $\displaystyle (k^\frac{8}{3}n^\frac{3}{2} )^\frac{-2}{3}$

How do i solve after this?

Like this? :$\displaystyle k^\frac{-16}{9}n^{-1}$ but how do i get the final answer $\displaystyle \frac{1}{nk}$ - Sep 28th 2008, 07:06 AMskeeter
had to fix my original post ... had the fraction flipped.

should be $\displaystyle x = k^{\frac{3}{2}}$

so, you should get ...

$\displaystyle (k^{\frac{3}{2}}n^{\frac{3}{2}})^{-\frac{2}{3}} = k^{-1}n^{-1} = \frac{1}{kn}$ - Sep 28th 2008, 08:06 AMshailen.sobhee
Find shortcuts - do not solve for x. See my steps how I got the answer.

Hope that helps. - Sep 28th 2008, 09:31 AMfabxx
Thanks to all three of you, I know how to do it now. Thanks guys. You guys were very helpful!!