[SOLVED] SAT math page 674 #19

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

Thanks in advance!! (:
• September 28th 2008, 04:42 AM
skeeter
this will get you started ...

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

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

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

now do the same to solve for y in terms of n ... then evaluate $(xy)^{-\frac{2}{3}}$
• September 28th 2008, 05:42 AM
fabxx
Quote:

Originally Posted by skeeter
this will get you started ...

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

How did you get this? Is this a rule? why multiply x with ^(-3/4)?

• September 28th 2008, 05:52 AM
Moo
Quote:

Originally Posted by fabxx
How did you get this? Is this a rule? why multiply x with ^(-3/4)?

$(a^b)^c=(a^c)^b=a^{bc}$

____________________________________________
other rules of exponents (among others) :
$a^b a^c=a^{b+c}$

$\frac{a^b}{a^c}=a^{b-c}$
• September 28th 2008, 06:05 AM
skeeter
Quote:

Originally Posted by fabxx
How did you get this? Is this a rule? why multiply x with ^(-3/4)?

my apologies ... I assumed you had some knowledge regarding the Laws of Exponents
• September 28th 2008, 06:42 AM
shailen.sobhee
Quote:

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

Thanks in advance!! (:

Basically,

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

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

and

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

From (1),

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

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

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

Multiply (2) and (3)

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

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

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

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

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

If i substitute I have $(k^\frac{8}{3}n^\frac{3}{2} )^\frac{-2}{3}$
How do i solve after this?
Like this? : $k^\frac{-16}{9}n^{-1}$ but how do i get the final answer $\frac{1}{nk}$
• September 28th 2008, 07:06 AM
skeeter
had to fix my original post ... had the fraction flipped.

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

so, you should get ...

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

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