Thank you to all those who bothered to reply. There is one que left which I think I did by how you guys showed me.
so,
ln(x-2) + lnx = 0
ln [x(x-2)] = 0
e^(ln x^2-2x) = e^0
x^2 - 2x = 0
is this correct? How should I proceed?
Thx!
2007-06-10 07:12:36 · 6 answers · asked by jill b 1 in Science & Mathematics ➔ Mathematics
You are almost there except that e^0 = 1 and not 0. Remember, any non-zero number raised to the power of zero is 1.
So you would have x^2 - 2x = 1
Now solve the above quadratic.
2007-06-10 07:17:33 · answer #1 · answered by cheeku_coolbuddy 1 · 2⤊ 0⤋
ln [ (x - 2).x ] = 0
(x - 2).x = e^0
(x - 2).x = 1
x² - 2x - 1 = 0
x = [ 2 ± â (4 + 4) ] / 2
x = [ 2 ± â 8 ] / 2
x = [ 2 ± 2â 2 ] / 2
x = 1 ± â 2
2007-06-10 15:02:00 · answer #2 · answered by Como 7 · 0⤊ 0⤋
x^2 - 2x -1 = 0
x = (-b ± â(b² - 4ac))/2a
x = (2 ± â(4 + 4))/2
x = (2 ± 2â2)/2
x = (1 ± â2)
.
2007-06-10 14:26:45 · answer #3 · answered by Robert L 7 · 0⤊ 0⤋
I agree with the other user answer
2007-06-10 14:20:47 · answer #4 · answered by Big D 2 · 1⤊ 0⤋
have you got paper 2 tomorro?
2007-06-10 14:24:33 · answer #5 · answered by Stan Darsh 3 · 0⤊ 0⤋
perhaps the internet is not the best place for maths.
2007-06-10 16:08:36 · answer #6 · answered by jgdyrh 2 · 0⤊ 0⤋