Yeah, I'm into logs and stuff now...and I don't get this problem:
Solve algebraically: ln(x+4) - ln(x-2) = lnx
I probably just didn't take good enough notes, but help would be appreciated.
Quote from: Silverhawk79 on December 16, 2007, 11:28:24 AM
Yeah, I'm into logs and stuff now...and I don't get this problem:
Solve algebraically: ln(x+4) - ln(x-2) = lnx
I probably just didn't take good enough notes, but help would be appreciated.
ln(x+4) - ln(x-2) = lnx
ln[(x+4)/(x-2)] = lnx
(x+4)/(x-2)=x
Quotient & Equal Property Rules, FTW. You can do the rest. ^_^
Quote from: NINNIN6 on December 16, 2007, 11:34:38 AM
Quote from: Silverhawk79 on December 16, 2007, 11:28:24 AM
Yeah, I'm into logs and stuff now...and I don't get this problem:
Solve algebraically: ln(x+4) - ln(x-2) = lnx
I probably just didn't take good enough notes, but help would be appreciated.
ln(x+4) - ln(x-2) = lnx
ln[(x+4)/(x-2)] = lnx
(x+4)/(x-2)=x
Quotient & Equal Property Rules, FTW. You can do the rest. ^_^
O_O I wish I could do that with my pre-algebra skills. -__-
Quote from: NINNIN6 on December 16, 2007, 11:34:38 AM
Quote from: Silverhawk79 on December 16, 2007, 11:28:24 AM
Yeah, I'm into logs and stuff now...and I don't get this problem:
Solve algebraically: ln(x+4) - ln(x-2) = lnx
I probably just didn't take good enough notes, but help would be appreciated.
ln(x+4) - ln(x-2) = lnx
ln[(x+4)/(x-2)] = lnx
(x+4)/(x-2)=x
Quotient & Equal Property Rules, FTW. You can do the rest. ^_^
No way, we JUST went over that.
My notes suck. >_>
Quote from: QuickKarel on December 16, 2007, 11:39:46 AM
Quote from: NINNIN6 on December 16, 2007, 11:34:38 AM
Quote from: Silverhawk79 on December 16, 2007, 11:28:24 AM
Yeah, I'm into logs and stuff now...and I don't get this problem:
Solve algebraically: ln(x+4) - ln(x-2) = lnx
I probably just didn't take good enough notes, but help would be appreciated.
ln(x+4) - ln(x-2) = lnx
ln[(x+4)/(x-2)] = lnx
(x+4)/(x-2)=x
Quotient & Equal Property Rules, FTW. You can do the rest. ^_^
O_O I wish I could do that with my pre-algebra skills. -__-
QFT, but wait, Im in Pre-algebra a year early :o.
Quote from: QuickKarel on December 16, 2007, 11:39:46 AM
Quote from: NINNIN6 on December 16, 2007, 11:34:38 AM
Quote from: Silverhawk79 on December 16, 2007, 11:28:24 AM
Yeah, I'm into logs and stuff now...and I don't get this problem:
Solve algebraically: ln(x+4) - ln(x-2) = lnx
I probably just didn't take good enough notes, but help would be appreciated.
ln(x+4) - ln(x-2) = lnx
ln[(x+4)/(x-2)] = lnx
(x+4)/(x-2)=x
Quotient & Equal Property Rules, FTW. You can do the rest. ^_^
O_O I wish I could do that with my pre-algebra skills. -__-
WHAT THE HELL? pre-algebra?
I took that poop in 5th grade and some in 6th.
I'm feeling dense tonight, and can't remember how to divide out fractions like that. -_-;
we did that earlier this year.
ln(x+4) - ln(x-2) = ln(x)
ln(x+4)/(x-2) = ln(x)
at this point, IIRC, you can de-log both sides.
x+4/x-2 = x
remember, any log added to another log is a multiplication. any log subtracted from another is a division.
ln is the same as log, except that log has a base of 10 and ln has a base of n.
Quote from: Jono2 on December 16, 2007, 05:50:04 PM
we did that earlier this year.
ln(x+4) - ln(x-2) = ln(x)
ln(x+4)/(x-2) = ln(x)
at this point, IIRC, you can de-log both sides.
x+4/x-2 = x
remember, any log added to another log is a multiplication. any log subtracted from another is a division.
ln is the same as log, except that log has a base of 10 and ln has a base of n.
I KNOW BUT I CAN'T REMEMBER WHAT TO DO TO FURTHER SIMPLIFY THE FRACTION
multiply it out...
x = x+4/x-2
x( x - 2) = x + 4
x2 - 2x = x + 4
x2 - 3x - 4 = 0
quadratic formula or factoring is the next step towards X.
but you don't need the quadratic formula, because it's
(x + 1) (x - 4)
and you can't have a negative log
so x is equal to 4.
Quote from: Jono2 on December 16, 2007, 06:19:34 PM
multiply it out...
x = x+4/x-2
x( x - 2) = x + 4
x2 - 2x = x + 4
x2 - 3x - 4 = 0
quadratic formula or factoring is the next step towards X.
but you don't need the quadratic formula, because it's
(x + 1) (x - 4)
and you can't have a negative log
so x is equal to 4.
Oh. I just multiplied it wrong.
Lulz.
Quote from: Pokesamrus on December 16, 2007, 04:49:25 PM
Quote from: QuickKarel on December 16, 2007, 11:39:46 AM
Quote from: NINNIN6 on December 16, 2007, 11:34:38 AM
Quote from: Silverhawk79 on December 16, 2007, 11:28:24 AM
Yeah, I'm into logs and stuff now...and I don't get this problem:
Solve algebraically: ln(x+4) - ln(x-2) = lnx
I probably just didn't take good enough notes, but help would be appreciated.
ln(x+4) - ln(x-2) = lnx
ln[(x+4)/(x-2)] = lnx
(x+4)/(x-2)=x
Quotient & Equal Property Rules, FTW. You can do the rest. ^_^
O_O I wish I could do that with my pre-algebra skills. -__-
WHAT THE HELL? pre-algebra?
I took that poop in 5th grade and some in 6th.
Me too. I'm in an Algebra 1 Honors class. :| And I'm in 7th grade.