Find x
2006-06-19 12:51:23 · 6 answers · asked by suxdikfd 1 in Science & Mathematics ➔ Mathematics
ln e^3 = 3
ln e^2 = 2
so... ln e^3 - ln 3^2 = 1
x = 1
2006-06-19 16:40:48 · answer #1 · answered by jimbob 6 · 0⤊ 0⤋
X=1
lne = 1
So 3 -2 = 1
2006-06-19 19:54:25 · answer #2 · answered by SOBITO 2 · 0⤊ 0⤋
2 ways to put this
ln(e^3) - ln(e^2) = x
3ln(e) - 2ln(e) = x
since the ln and e^y cancel each other out, this leaves you with
3 - 2 = x
x = 1
or you can do it like this
ln(e^3) - ln(e^2) = x
ln((e^3)/(e^2)) = x
ln(e^(3 - 2)) = x
ln(e^1) = x
ln(e) = x
x = 1
2006-06-19 20:12:57 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
ln(e^3) - ln(e^2) = x
ln( e^3 / e^2 ) = x
ln e = x
x = 1
OR
3ln e - 2ln e = x
ln e = 1
3-2 = x = 1
2006-06-19 19:54:48 · answer #4 · answered by the redcuber 6 · 0⤊ 0⤋
ln x = log_e x
so
ln e^c = log_e e^c
but
log_b b^n = n,
so
ln e^c = c
ln e^3 - ln e^2 = x
so
3 - 2 = x
x = 1
^_^
2006-06-20 06:41:47 · answer #5 · answered by kevin! 5 · 0⤊ 0⤋
? 1!
2006-06-19 19:56:14 · answer #6 · answered by wayladuley 3 · 0⤊ 0⤋