Can u solve this equation?

0 ⤊

ln (1+x) = 1+ lnx

Thanks

2007-09-27 04:36:50 · 5 answers · asked by Mystic healer 4 in Science & Mathematics ➔ Mathematics

yes I can solve it.

2007-09-27 04:39:33 · answer #1 · answered by Anonymous · 0⤊ 2⤋

ln (1+x) = 1+ lnx
= ln (1 + x) - ln x = 1
=> ln [ (1 + x) / x ] = 1
=> (1 + x ) / x = e
=> 1/x + 1 = e
=> 1/x = e - 1
=> x = 1 / (e - 1)

2007-09-27 11:45:34 · answer #2 · answered by Madhukar 7 · 0⤊ 0⤋

1/(e - 1)

2007-09-27 11:40:11 · answer #3 · answered by PMP 5 · 0⤊ 0⤋

e^ln(1+x) = e^(1+lnx)
x+1 = e^1e^lnx = ex
ex-x =1
x(e-1)=1
x = 1/(e-1) approximately =.5819767069

2007-09-27 12:01:25 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋

ln(1+x)=ln(e)+ln(x)
ln(1+x)=ln(e*x)
1+x=e*x
(e-1)*x=1
x= 1/(e-1)

2007-09-27 11:46:23 · answer #5 · answered by Andy S 6 · 0⤊ 0⤋