Cloudy's Cafe-a conversation zone, 5-9-02

[Sazerac]

Originally posted by Cloudbringer: Poor Elif! There must be a timestop spell handy so you can have your quality killing time in the game! I'm doing ok, Bahamut, but my co-worker went home, sick, a few hours ago and it's busy here! Today is the 'reading day' which is the day between the last day of classes and the first day of final exams so all the students and faculty have been in all day!

Hi, Cloudy! It's been a madhouse here at IBS this week, too...it's the first week of our new session and trying to get all the new student accounts for Word and Excel and Powerpoint...whew! Plus my super is being called away to do Army Reserve Duty for the next week. Poor lady, she's due to retire as a Master Sergeant with full honors in the reserve on May 22, and they're calling her in for the week of the 13th to the 17th! Pah. I told her they're getting all the "fo' free" out of her they can, and she agrees, but she's got to do it, or they can recall her after she's retired. She's wanting a CLEAN retirement, and I don't blame her. But anyway, I've got to cover her TCP/IP protocols class for her next week...more work for Papa here! [img]tongue.gif[/img] At least it's during the day, and I don't have to be preempted for night. (whew!)

-Saz

[Sazerac]

Originally posted by Neb: Hm, alright, it's just that we haven't worked that much with ln and are in the middle of working with log.

Oh! Well, you can do it with log, too...just substitute it for ln in Azred's formula. It all cancels out, regardless of what base you use. [img]smile.gif[/img]

Cheers,
-Saz

[Azred]

Sazerac is correct; without ln, the best method is iteration (but iteration can be time-consuming without electronic assistance).

@Neb: log x = (ln x) / 2.30258 [img]graemlins/petard.gif[/img]

[Neb]

Originally posted by Sazerac: quote: Originally posted by Neb: Hm, alright, it's just that we haven't worked that much with ln and are in the middle of working with log.

Oh! Well, you can do it with log, too...just substitute it for ln in Azred's formula. It all cancels out, regardless of what base you use. [img]smile.gif[/img]

Cheers,
-Saz[/QUOTE]Goody! Now how much more maths can I convince you guys to do for me?

1) log(x+1) + log(x-1) = 0

2) log((x^2)-1) - log(x+1)-1 = 0

[Cloudbringer]

HA! I see the math wizzes have come out of the woodwork!

Hey Elif, I hear you! So you sleep what? 4 HOURS A NIGHT?! Mercy!, that capslock thing is contagious!

HAHAHA..my keyboard is squeaking! LOLOL! [img]graemlins/biglaugh.gif[/img]

[Azred]

Originally posted by Neb: Goody! Now how much more maths can I convince you guys to do for me? 1) log(x+1) + log(x-1) = 0 2) log((x^2)-1) - log(x+1)-1 = 0

For you? None. [img]graemlins/firedevil.gif[/img] However, for the first one just move one term and exponentiate both sides. For the second, recall that x^2 - 1 = (x+1)(x-1) and that log (ab) = log a + log b. Simple, yes? [img]graemlins/petard.gif[/img]

[ 05-09-2002, 03:38 PM: Message edited by: Azred ]

[Azred]

Originally posted by Cloudbringer: HA! I see the math wizzes have come out of the woodwork!

Thank goodness you didn't include an "h" in "wizzes"! [img]graemlins/beigesmilewinkgrin.gif[/img]

[Sazerac]

Originally posted by Neb: quote: Originally posted by Sazerac: quote: Originally posted by Neb: Hm, alright, it's just that we haven't worked that much with ln and are in the middle of working with log.

Oh! Well, you can do it with log, too...just substitute it for ln in Azred's formula. It all cancels out, regardless of what base you use. [img]smile.gif[/img]

Cheers,
-Saz[/QUOTE]Goody! Now how much more maths can I convince you guys to do for me?

1) log(x+1) + log(x-1) = 0

2) log((x^2)-1) - log(x+1)-1 = 0[/QUOTE]On logs, when adding them, you multiply their arguments. Like: Log A + log B = Log AB. When subtracting logs, you divide their arguments. Like: Log A - Log B = Log (A/B).

So for your first one: log(x+1) + log(x-1) = log((x+1)(x-1) or log(x^2 - 1). Set that equal to zero, take the antilog of both sides, and solve for X.

(I'd do it for you, but A) you need to learn to do it yourself, and B) some yayhoo deleted the Windows calculator off this computer, the swine! [img]graemlins/1pissed.gif[/img] ) Ah well...hope that helps!

-Saz

[Neb]

Originally posted by Sazerac: quote: Originally posted by Neb: Goody! Now how much more maths can I convince you guys to do for me? 1) log(x+1) + log(x-1) = 0 2) log((x^2)-1) - log(x+1)-1 = 0

On logs, when adding them, you multiply their arguments. Like: Log A + log B = Log AB. When subtracting logs, you divide their arguments. Like: Log A - Log B = Log (A/B).

So for your first one: log(x+1) + log(x-1) = log((x+1)(x-1) or log(x^2 - 1). Set that equal to zero, take the antilog of both sides, and solve for X.

(I'd do it for you, but A) you need to learn to do it yourself, and B) some yayhoo deleted the Windows calculator off this computer, the swine! [img]graemlins/1pissed.gif[/img] ) Ah well...hope that helps!

-Saz[/QUOTE]Heh, I already know how to do it. I'm just too damn lazy to do it this time [img]tongue.gif[/img] And, ah, doing it this way allows me to both surf the web and do my homework at the same time

Oh and, they're two separate things. That's why the 1) and the 2).

[ 05-09-2002, 03:48 PM: Message edited by: Neb ]

[Cloudbringer]

Hi Sazzie! Yes if your supervisor has duty the last month of her service she has to do it! LOL They really do get all they can out of you! I'm sure I'll be happy to kill stuff in IWD now! This place is a zoo this week! I am just looking forward to a fast food dinner, some IWD tromping about and killing of things and talking with my sweetie before I get to bed.

Cloudykins