2006-11-09 05:14:21 · 9 answers · asked by Dennis B 1 in Science & Mathematics ➔ Mathematics
ln[x+5]-ln3=ln[x-3]
ln[x+5]-ln3-ln[x-3]=0
ln[(x+5)/3(x-3)=0
[x+5]/3[x-3]=1
3[x-3]=x+5
3x-9=x+5
2x=14
x=7
verify
ln12-ln3=ln4
ln12/3*4
=ln1
=0
ok
2006-11-09 05:24:06 · answer #1 · answered by openpsychy 6 · 0⤊ 0⤋
I guess the equation is ln(x+5) - ln(3) = ln(x - 3)
First must be x>3, otherwise ln(x-3) does not exist.
ln(x+5) - ln(3) = ln(x - 3)
ln{(x+5)/3} = ln(x - 3) or
(x + 5) / 3 = x - 3
x + 5 = 3x - 9
14 = 2x
x = 7 (and that is > 3) So the solution is x = 7
Th
2006-11-09 05:39:28 · answer #2 · answered by Thermo 6 · 0⤊ 0⤋
ln(x+5)-ln(3)=ln(x - 3)
ln (x + 5)/3 = ln (x - 3)
x+5 = 3*(x-3)
x + 5 = 3x - 9
5 + 9 = 2x
x = 7
2006-11-09 05:42:02 · answer #3 · answered by Dr. J. 6 · 0⤊ 0⤋
ln(x+5)/3=ln(x-3)
=>(x+5)/3=(x-3)
=>(x+5)=3(x-3)
=>x+5=3x-9
=.>2x=14 andso x=7
2006-11-09 05:17:58 · answer #4 · answered by raj 7 · 1⤊ 0⤋
(x+5)/3 = x-3
or,
x+5 = 3x -9
2x = 14, x=7.
2006-11-09 05:18:37 · answer #5 · answered by Sumit B 1 · 1⤊ 0⤋
Are those supposed to be natural logs? ("ln" instead of "1n")
2006-11-09 05:18:04 · answer #6 · answered by Dave 6 · 0⤊ 1⤋
I've seen people use the "L" key to type a "1"... but the other way around???!?
2006-11-09 05:58:26 · answer #7 · answered by Anonymous · 0⤊ 0⤋
I can't because my brain just froze from eating ice cream too fast.
2006-11-09 05:22:55 · answer #8 · answered by Anonymous · 0⤊ 1⤋
nx+5n-3n=nx-3n
nx + 5n = nx (add 3n to each side
5n = 0 (subtract nx from each side)
n=0
2006-11-09 05:19:38 · answer #9 · answered by SHAWN G 3 · 0⤊ 2⤋