If lnx+ln(x–4)=ln2x?

0 ⤊

If lnx+ln(x–4)=ln2x?

If lnx+ln(x–4)=ln2x
then x=?

2007-10-30 19:01:59 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

3 answers

(x) (x - 4) = 2x
x² - 4x = 2x
x² - 6x = 0
x (x - 6) = 0
x = 0 , x = 6
Accept value of x = 6

2007-10-31 03:18:41 · answer #1 · answered by Como 7 · 0⤊ 0⤋

ln(x) + ln(x-4) = ln(2x)
ln[ x(x-4) ] = ln 2x
x(x-4) = 2x
x² - 4x = 2x
x² - 6x = 0
x(x-6) = 0
x = 0 or x = 6,
but x = 0 leads to ln(0-4) = ln(-4), no solution if x real, so
x = 6

2007-10-30 19:07:01 · answer #2 · answered by Philo 7 · 0⤊ 1⤋

lnx+ln(x-4)=ln2x
ln(x(x-4))=ln2x
x(x-4)=2x
x^2-4x-2x=0
x^2-6x=0
x(x-6)=0
x=0 or x=6, however x=0 is rejected.
therefore x=6

2007-10-30 19:09:26 · answer #3 · answered by someone else 7 · 0⤊ 1⤋