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
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∙ 12y agoThink 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
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
3m^2 - 9m = 0 factor out 3m 3m(m - 3) = 0 m = 0 -------- or m = 3 --------
3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)
e2x=ex^2 basically means that x2=2x, in which case x2-2x=0, x = 0, 2. I don't think that's what the question meant. It could mean: e2x=(ex)2 . Which comes from one of the rules of exponents. Basically, look at it this way: Take the natural log of both sides: ln e2x= ln(ex)2 From rules of logs: (2x) ln e = (2) ln ex 2x ln e = (2) (x) ln e 2x = (2)(x)
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
-6
Nope - 9 - 0 = 9 !
2x1=1 so 1-Y=0 so Y=0, was this really a question?
If it is 20(3-3) it is equal to 0. 20X3-3= 57
0
-16 minus -16 is 0.
-10 minus -10 is 0.