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lolz pre-calc halp

Started by Silverhawk79, December 16, 2007, 11:28:24 AM

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Silverhawk79

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.

Rius

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. ^_^

QuickKarel

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. -__-


Silverhawk79

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. >_>

BOSSNIG

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.
IF U WERE KILLED TOMORROW, I WOULDNT GO 2 UR FUNERAL CUZ ID B N JAIL 4 KILLIN DA MOTHA FUKER THAT KILLED U!
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.....), ---.(_(__) /
....// (..) ), ----"
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..//___//
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WE TRUE HOMIES
WE RIDE TOGETHER
WE DIE TOGETHER

Pokesamrus

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.

Silverhawk79

I'm feeling dense tonight, and can't remember how to divide out fractions like that. -_-;

Jono2

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: LinkXLR on January 30, 2008, 09:10:54 PM
Quote from: famy on January 30, 2008, 08:36:30 PM
is big willy unleashed a will smith game

...I'm not even gonna touch this one.

SteamID: Lazylen

Silverhawk79

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

Jono2

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: LinkXLR on January 30, 2008, 09:10:54 PM
Quote from: famy on January 30, 2008, 08:36:30 PM
is big willy unleashed a will smith game

...I'm not even gonna touch this one.

SteamID: Lazylen

Silverhawk79

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.

Daft Pink

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.