so, if 2 minus Ln times 3 minus x equals 0, then 2 minus Ln times 3 equals x,
therefore 2 minus Ln equals x divided by three, so Ln + X/3 = 2
therefore, (Ln + [X/3]) = 1
Think of ln 1 as "e to what power will given me 1." Anything to the zero power will give you 1. So, ln 1 = 0, and 0/2 = 0
3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)
3m^2 - 9m = 0 factor out 3m 3m(m - 3) = 0 m = 0 -------- or m = 3 --------
log3(x) + log9(x^3)=0 for any t >0 logt(x) = ln(x)/ln(t) so log9(x) = ln(x) / ln (9) = ln (x) / ( 2 * ln 3) = log3(x) /2 or log3(x) = 2 log9(x) log9(x^n) = n * log9(x) So log3(x) + log9(x^3) = log3(x) + 6 log3(x) = 7 log3(x) 7 log3(x) = 0 => log3(x) = 0 => x = 1
d/dx[ln(10)] = 0
ln 1 = 0 e0=1
Minus 1 times 0 plus 3 divided by 3 is equal to 1
It is equal to 0
=1
0
1
-6
Nope - 9 - 0 = 9 !
0
2x1=1 so 1-Y=0 so Y=0, was this really a question?
-10 minus -10 is 0.
-16 minus -16 is 0.