Recent content by hancyu

so the domain of the 1st one is x=>0 ? because log can never be zero or negative? i still can't get the inverse tho...

also, is the inverse of f(x) = 2x−1 + 3 log2 (x/3) +1 = y f(x) = log2/3(x − 2) − 4 (2/3)x+4 + 2 = y are these correct?

Homework Statement f(x) = log2 x + 3 2 log2 x − 1 how do i find the inverse of this? how do i find the range and domain of a log function? Homework Equationsf(x) = log2 x + 3 2 log2 x − 1 is equal to f(x) = log2 x + 3 - log2 (x − 1)2 D of f(x) = R of f-1(x) The...

is the answer to the problem y= ln(x/2) ?

thank you very much!:biggrin:

ahhh! 1/3 (23) + 64/3 = 29 ? so the sum is 29

oh. I am really sorry. 1/3(a1...a16) + 64/3 = 23 so the sum is 5 :D

ok... 1/3(a1+a2+a3+a4+a5+a6+a7+a8+a9+a10+a11+a12+a13+a14+a15+a16) + 64/3 = 0 so... the sum is 64?

what is a' + b' + … + p' ? how do i add it? there are many different variables? is it 1/3(a...p) + 64/3 ?

im sorry. i really don't get this. what's a' ?

f(3(a/3 + 4/3) - 4) = 0 ?

how do i get the sum...there are 16 unknown variables...a,b,c,d,e...p i think i got the answer on the ellipse problem... length of each side is 24/5 area is 576/25.

f(a/3 + 4/3) = 0 ?

a/3 ? so the zero should be... a/3 +4 , b/3 +4...

the zeros of f(3x) are 3x-a, 3x-b, 3x-c, 3x-d... then the zeros of f(3x-4) are 3x-4a, 3x-4b... is that correct?