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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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Q: What is the answer to 2 minus Ln times 3 minus x equal 0?

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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

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)

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