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Old 11-04-08, 07:19 PM   #1
phantomcow2
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Is this logarithm equation properly condensed?

ln(e+1) - ln(e^2) + ln(e) =

ln(e*1) = ln(e)

ln(e)/ln(e^2)(e) Divide because of the subtraction.

but ln(e) = 1, and ln e^2 * e = e^3. Lne^3 = 3.

Answer = 1/3
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Old 11-04-08, 07:30 PM   #2
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if you foo egg heads are gonna go at it tonight don't just stand there - pls count us some votes or something!
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Old 11-04-08, 07:47 PM   #3
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no...?
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Old 11-04-08, 07:51 PM   #4
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I am working on another problem right now that is far more pressing. I am trying to take a pile of about 150 red poker chips and turn them into more than 270, all using only the will of my mind...

It's proving to be difficult.

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Old 11-04-08, 07:53 PM   #5
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I wrote an amusing solution to Phantom's question in the margin of Bike Forums. But now I'm feeling...argh...[thud]
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Old 11-04-08, 07:54 PM   #6
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well, so long as you are using your time productively

seriously, you guys with your physics and organic chem threads are making me feel as stupid as I really am.

I am thinking about taking some night classes in math/science. I'm one of those humanities people, so it'll be like teaching Gary Coleman to slam dunk. But life needs a challenge, huh?
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Old 11-04-08, 07:57 PM   #7
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I'm having a hard time understanding what you're doing...

Going kinda along your lines:

ln(e + 1) - ln(e^2) + ln(e) =
ln((e + 1)/e^2) + ln(e) =
ln(((e + 1)/e^2)*e) =
ln((e + 1)/e)=
ln(e + 1) - ln(e) =
ln(e + 1) - 1


Or more simply:


ln(e + 1) - ln(e^2) + ln(e) =
ln(e + 1) - 2*ln(e) + ln(e) =
ln(e + 1) - 2*1 + 1=
ln(e + 1) - 1
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Old 11-04-08, 08:03 PM   #8
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and there's why this electronic engineer wannabe settled for computer analyst nighshifter...
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Old 11-04-08, 08:14 PM   #9
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Remember that:



and

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Old 11-04-08, 08:21 PM   #10
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Quote:
Originally Posted by phantomcow2 View Post
ln(e+1) - ln(e^2) + ln(e) =

ln(e*1) = ln(e)

ln(e)/ln(e^2)(e) Divide because of the subtraction.

but ln(e) = 1, and ln e^2 * e = e^3. Lne^3 = 3.

Answer = 1/3
Attempting to go by your logic, I think where you may have gone wrong was assuming that this was an equation, i.e. you assumed that there was something that it related to. The problem was asking you to reduce the expression, so there was nothing to move that ln(e) statement to.
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Old 11-04-08, 08:21 PM   #11
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oh yeah, I remember now
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Old 11-05-08, 08:59 AM   #12
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Stay away from condensed expressions, dude. I only use freshly-squeezed expressions.
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Old 11-05-08, 10:52 AM   #13
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Make sure it doesn't have concentraet.
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Old 11-05-08, 11:14 AM   #14
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No, you need to concentrate in order to solve these problems.
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Old 11-05-08, 11:21 AM   #15
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Quote:
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Remember that:



and

ln e^a = a

Last edited by MrCrassic; 11-05-08 at 11:24 AM.
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Old 11-05-08, 11:21 AM   #16
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E = sightsee MC
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Old 11-05-08, 11:24 AM   #17
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Quote:
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ln e^a = a
That's true, but I was demonstrating the principle.
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Old 11-05-08, 11:26 AM   #18
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Quote:
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That's true, but I was demonstrating the principle.
yeah, I got it, I just like the shorthand version better. Much less messy...
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Old 11-05-08, 11:31 AM   #19
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Old 11-05-08, 12:07 PM   #20
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Thank John Napier you have them as a tool in the first place.
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- it's pretty well established that Hitler was an *******.
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Old 11-05-08, 03:07 PM   #21
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Thanks to those who contributed with helpful comments.
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Old 11-05-08, 03:14 PM   #22
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you're welcome
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Old 11-05-08, 03:18 PM   #23
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Old 11-05-08, 03:24 PM   #24
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We need more of these in schools:






And fewer of these:

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Old 11-05-08, 04:22 PM   #25
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Quote:
Originally Posted by apricissimus View Post
We need more of these in schools:

<snipped image>

And fewer of these:

<snipped image>
Agreed.

When I was in high school, my physics teacher had a supply of slide rules plus instruction booklets that were handed out to any student who forgot their calculators (this was when all the cool kids had TI-86s) on exam days. It then became a bit of a machismo contest to see who could get the best marks using the slide rules. By the end of the course, we realized that the graphing calculators aren't nearly as essential as we once thought.
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