Ok so I have this: So pretty much, x^-n = 1/y^2 is a, b, c or d. Thank you guys sooooo much PS. Fack I typoed the topic name XD Thats what happens when you have someone trying to get attention around you all the time .

Schools over for me and I already forgot all of this, but I think its ... nevermind I forgot how to do these EDIT: i did it, got a Let me upload work.... good luck reading it

Has being so long since school but if I remember correctly: Log(base) n = e :: base^e = n so then the most acurrated answer is C because Log(x) y = -2n :: x^(-2n) = y How to pass the exponent (2) to the other side?, I don't remember but that seems the most logic answer to me. I hope that can help you. EDIT: Wait, now that I think about it the answer also could be A because: Log(x) y = n/2 :: x^(n/2) = y Where is the (-)?, I have no idea but now you have two posibles answers out of four.

Linki, I got it but I don't understand how raising to the -1 power at the end would make n positive? Wouldn't it be 1/-n?

A is right, you simplify -1/n by factoring out -1.. it's just one of those fundamental principles about roots cannot be negative..

A.. pretty easy How about solving this one Log y X + 2Log x Y = 4 XY=9 Note : the small y and x mean that they are the base

Here are two rules: log(ab) = log(a) + log(b) log(a^2) = 2log(a) (can be derived from the formula above, use another a in b's place) x^(-n) = 1 / (y^2) This can be re-written as x^(-n) * y^2 = 1 1) Take log base x on both sides Use the first rule I wrote above, and log(1) = 0 for any base of log. logx (x^(-n)) + logx (y^2) = 0 logx (x^(-n)) is just -n, logx (y^2) can be re-written as 2 logx(y) (Look at the second rule above) So, -n + 2 logx(y) = 0 add n to both sides and divide both sides by 2 logx(y) = n / 2 2) Take log base y on both sides logy(x^(-n)) + logy(y^2) = 0 -n logy(x) + 2 = 0 subtract 2 from both sides and divide both sides by -n logy(x) = 2/n Edit: Wait, regardless of which one you do first, you have to do log base x anyway. =P