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!! (:
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 $\displaystyle (xy)^{-\frac{2}{3}}$
my apologies ... I assumed you had some knowledge regarding the Laws of Exponents
Basically,
If its SATs, AS A RULE, forget about complex rules.
$\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