[beary605]

i'm not sure how math methods works (it could be about calculation efficiency or something) but any solution should be fine, so long as all the steps are right. though i guess a general rule of thumb is to always try to work with small numbers and as few things as possible, so there's less chance of messing something up (and not noticing it later)

first one: i'd say flip the indices first, because then the x and y terms can combine and there'll be less terms to exponentiate
2x^2 * (x^-1*y^2/2xy^-1)^3
= 2x^2 * (y^2*y^1/2x*x^1)^3
= 2x^2 * (y^3/2x^2)^3
= 2x^2 * y^9 / 8x^6
= y^9 / 4x^4

second one: the algebra's a little off but expanding it out is what i'd do
(e^2x + e^-x)*(e^2x - e^-x)
= e^2x*e^2x - e^2x*e^-x + e^-x*e^2x - e^-x*e^-x
= e^4x - e^x + e^x - e^-2x
= e^4x - e^-2x

alternatively the product's of the form (a+b)(a-b) = a^2-b^2 [where a=e^2x, b=e^-x] and you can multiply it like this:
(e^2x + e^-x) * (e^2x - e^-x) = (e^2x)^2 - (e^-x)^2 = e^4x - e^-2x
but both methods are equally good

third one: you got it

fourth one: yeah i'd divide cause it'd keep the x indice small
3 ln(2x) - ln(4x^4) + 2 ln(3x^2)
= ln[(2x)^3)] - ln(4x^4) + ln[(3x^2)^2)]
= ln(8x^3) - ln(4x^4) + ln(9x^4)
= ln(2x^-1) + ln(9x^4)
= ln(18x^3)

there's a little trip-up with multiplying too because of the negative sign on ln(4x^4): if you multiply you'd actually be calculating ln(8x^3) - [ln(4x^4) + ln(9x^4)] = ln(8x^3) - ln(4x^4) - ln(9x^4). i guess you could imagine brackets around the group you want to divide/multiply, and see if that's equivalent to the equation you're trying to simplify (or avoiding negatives in general)

if you wanna double-check your answer you can use wolframalpha or another algebra tool (some of them have step-by-step generators too, though the method they use might not be the prettiest)

anyways, hope that helps

edit: congrats on 666 posts :P