<p>I know there have been threads about this problem but I'm still unclear:
(x)^ (-4/3)= (k)^ (-2) and (y) ^ ( 4/3)= n^2, what is (xy) ^ ( -2/3) in terms of n and k?
a) 1/ nk
b) n/k
c) k/n
d) nk
e) 1</p>
<p>So here's my solution:
n^2= y^ (4/3) => n= ((y)^2/3)^2 => n= y ^(2/3)</p>
<p>k^-2= x^-(4/3) => k= (x^ (2/3))^2 => k= x^(-2/3)</p>
<p>(xy)^ (-2/3)
= x^ (2/3) y ^ (-2/3)
= x 1/n
= k/n</p>
<p>So the answer is C. I don't know why BB says A</p>
<p>you can manipulate the first equation to be k^2 = x^4/3</p>
<p>you also have y^4/3 = n^2</p>
<p>so (xy)^4/3 = (kn)^2</p>
<p>square root both sides:
(xy)^2/3 = kn</p>
<p>take reciprocals:
(xy)^-2/3 = 1/kn (Answer A–you error occurs in your second derivation 1/k = x^(-2/3))</p>
<p>By the way, what does 19/674 mean, and what # was this in BB?</p>
<p>Number 19 page 674, # is the symbol for number. I appreciate your way of solving, cadillac, but can you solve it my way so that I know what’s wrong with my logic?</p>
<p>Yes, as I said before, your error occurs in your second derivation; otherwise, your method is fine.</p>
<p>1/k = x^(-2/3)</p>
<p>I think your problem is that you are trying to do too many steps at once. You wrote:</p>
<p>k^-2= x^-(4/3) => k= (x^ (2/3))^2 => k= x^(-2/3)</p>
<p>Stretch it out:</p>
<p>k^-2=x^(-4/3)=>k^2=x^(4/3)=>k<em>k=x^(2/3)</em>x^(2/3)=>k=x^(2/3)=>1/k=x^(-2/3)</p>
<p>Your line has two errors:</p>
<p>k= (x^ (2/3))^2 is not true k= (x^ (4/3))^(1/2)=> k=x^(2/3)</p>
<p>and the second one as stated by cadillac.</p>
<p>That was a typo. ohlwiler. But a thing I dont understand is if we don’t multiply by ^-1, and keep the negative ^ ( I forgot what it’s called lol), can you please show your work if we do it like that?</p>