| 00:12.45 | LeviHome | Quiet today. |
| 00:48.55 | mheath | This is off topic, but would anyone here be able to help me with some math homework? (Logorithms..) |
| 01:06.12 | *** join/#utah mheath_ (~Michael@c-67-161-213-34.client.comcast.net) |
| 01:21.04 | *** join/#utah carruth (carruth@sverige.freeshell.org) |
| 01:38.27 | *** join/#utah decriptor (~decriptor@166.70.98.58) |
| 01:38.42 | decriptor | hey all |
| 01:39.02 | decriptor | anyone around willing to help me with a bind 9 issue |
| 01:45.00 | *** part/#utah tensai (~tensai@kite.zmonkey.org) |
| 01:47.28 | LeviHome | mheath: Go ahead and tell us your problem. I can't guarantee I'll be of any help, though. |
| 01:49.31 | mheath_ | Heh, actually, when I asked that I was on the console in a very bad IRC program that is very hard to use in multiple channels.... |
| 01:49.37 | mheath_ | Meant to ask in another channel |
| 01:50.11 | mheath_ | I have several questions that have popped up as I've been going through this assignment |
| 01:50.25 | mheath_ | 2 * log (x+3) = log ((x+3)^2) correct? |
| 01:50.42 | decriptor | yeah I think so |
| 01:51.26 | mheath_ | That doesn't make sense, though........because with the rules of logs, that would mean that: |
| 01:52.03 | mheath_ | 2 * log (x+3) = log (x+3) + log(x+3) |
| 01:52.11 | mheath_ | Or do logorithms have that relationship? |
| 01:52.41 | decriptor | no |
| 01:52.52 | goozbach | this is neato |
| 01:52.55 | decriptor | I'm pretty sure that your first one is right |
| 01:53.44 | mheath_ | I hate the way my brain works. |
| 01:53.49 | goozbach | sitting in a boring meeting, ssh'ed into an IRC session |
| 01:54.11 | mheath_ | I'll sit and try to figure something out, not being able to make a bit of sense of it...it just doesn't fit, doesn't work, nothing makes sense.. |
| 01:54.25 | mheath_ | Then, at some random moment, it all clicks into place, makes perfect sense |
| 01:54.33 | decriptor | maybe this will help a little http://www.mathsdirect.co.uk/pure/purtutloglaw.htm |
| 01:54.43 | goozbach | (on my treo) |
| 01:54.52 | mheath_ | descriptor: I figured it out, thanks though :) |
| 01:54.58 | decriptor | ok |
| 01:55.06 | mheath_ | goozbach, what kind of internet connectivity do treo's have? |
| 01:55.07 | decriptor | anyone here good a bind 9 stuff |
| 01:55.26 | goozbach | its not too shabby |
| 01:55.51 | mheath_ | No, I mean....is it just normal WiFi? From what you've been saying lately, sounds like you get internet access almost everywhere |
| 01:56.38 | goozbach | the laentcy is horribe like 1 sec per char in this ssh client |
| 01:56.48 | *** join/#utah bonez39 (~aint@c-67-166-77-14.client.comcast.net) |
| 01:56.48 | goozbach | it's over pcs |
| 01:56.55 | mheath_ | Ah! Awesome |
| 01:57.00 | mheath_ | You make me jealous :) |
| 01:57.07 | goozbach | yeah! |
| 01:57.40 | goozbach | qme leaves |
| 01:57.41 | LeviHome | goozbach: What ssh program do you use? |
| 02:01.54 | Newsome | Is the connectivity expensive? |
| 02:12.43 | LeviHome | No, it's the regular Sprint PCS Vision service. |
| 02:13.01 | findlay | mheath_: I usually check wikipedia.org for math stuff, rules, theorems, and stuff, or horrors, mathworld |
| 02:21.40 | *** join/#utah tensai (~tensai@kite.zmonkey.org) |
| 02:42.38 | *** join/#utah ECS (~ecs@ecs-dell.byu.edu) |
| 02:43.31 | findlay | ~help |
| 02:44.49 | findlay | ~wikipedia logarithm |
| 02:45.00 | goozbach | pssh |
| 02:45.30 | findlay | ~hex logarithm |
| 02:45.30 | ibot | logarithm is 6C 6F 67 61 72 69 74 68 6D |
| 02:45.41 | findlay | darn, I was hoping for a spell |
| 02:46.13 | tensai | ~spell logarithm |
| 02:46.14 | ibot | 'logarithm' may be spelled correctly |
| 02:46.35 | tensai | oh, wait. now I get it. |
| 02:47.37 | findlay | no, ibot, it may not be |
| 02:48.13 | findlay | ~math logarithm |
| 02:51.48 | goozbach | $15/month |
| 02:52.05 | goozbach | unlimited pcs |
| 02:52.11 | findlay | cool |
| 02:52.48 | goozbach | the phone is mucho expensivo however |
| 02:53.33 | mheath_ | Anyone have any idea how I'd solve something like: |
| 02:53.40 | mheath_ | 1/49 = 7^(2x+3) |
| 02:54.05 | goozbach | log |
| 02:54.11 | mheath_ | I know that's equal to ln(1/4) = (2x+3) (ln 7) |
| 02:54.24 | mheath_ | But I'm not sure where to go from there |
| 02:54.33 | findlay | isolate x? |
| 02:54.38 | goozbach | yeah natural log |
| 02:54.50 | mheath_ | well... |
| 02:54.50 | mheath_ | sl |
| 02:54.58 | mheath_ | er |
| 02:55.11 | mheath_ | 2x+3 = ln(1/4) |
| 02:55.17 | findlay | x = (ln(1/4)/ln 7 + 3)/2 |
| 02:55.18 | mheath_ | Argh, I hate this keyboard |
| 02:55.29 | findlay | x = (ln(1/4)/ln 7 - 3)/2 |
| 02:55.57 | mheath_ | Sorry, this keyboard isn't the one I usually use and the enter keys a different shape.. |
| 02:56.10 | mheath_ | findlay, is that the best it's possible to get it to? |
| 02:56.13 | findlay | I hate that |
| 02:56.13 | mheath_ | The simplest, I mean |
| 02:56.22 | findlay | I don't think so |
| 02:56.45 | findlay | I think there's some property of logarithms to translate fractions |
| 02:56.45 | mheath_ | Yeah, I'm thinking the same. I have no problem getting it to that point, I'm just not sure how to get it from there.. |
| 02:56.52 | findlay | but I don't remember it |
| 02:57.26 | mheath_ | Well, I know rules that apply to ln ( x / y), but none that apply to ln (x) / ln (y) |
| 02:57.38 | findlay | especially those wierd MS keyboards, with the special function keys block all scrambled |