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Spock_Logic147
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Joined on: 06 Jun 2008
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 Posted: Mon Mar 01, 2010 8:48 pm    Post subject: Accumulating Equation In this game, we will continuously create an equation in which you have to solve for x. How it works: I will start off the game with something. The poster after me can change or add a value or symbol to the equation, but must first post the current value of x (as determined by the equation of the poster above). The equation continues to build until someone either gets the wrong value for x (must be watched by other posters) or they make an equation that cannot be solved. The person who has wronged then receives 2 SB. A new equation can be created by whoever posts next... Rules: -All numbers must be in base 10. -You may only add or change one operand or one operator in your equation mod -Calcnoobs, let's try to not immediately kill the game... -Yes...you must list both values for x in cases like square roots For your fancy math symbols, use this EDIT: Had to shorten URL START: _________________Here's the thing...
JMickle
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 Posted: Mon Mar 01, 2010 10:58 pm    Post subject: x=0.5 [img]http://www.sitmo.com/gg/latex/latex2png.2.php?z=175&eq=e^\frac{2}{x}%3D4[/img]_________________Hello I am a regular person!
Spock_Logic147
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 Posted: Tue Mar 02, 2010 3:08 am    Post subject: Apparently the linked image does not work..that sucks. Guess imageshack is needed... Here was Jmick's equation: cmon guys..this could get interesting_________________Here's the thing...
psy_wombats
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 Posted: Tue Mar 02, 2010 3:44 am    Post subject: x = 1/ln(2) _________________When Wombats Strike!
JMickle
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 Posted: Tue Mar 02, 2010 4:57 pm    Post subject: x=ln(4) e^(sin(2x)) = 4^x_________________Hello I am a regular person!
psy_wombats
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 Posted: Tue Mar 02, 2010 5:07 pm    Post subject: Ha, whoops, I accidentally changed 2 things... That was supposed to be 2/x instead of 2x, but in any case, pretty sure the only solution to the printed equation was x=0. ln(4) yields 16 for the left hand side and something like 7 for the right. So we'll go back to what I intended: e^(2/x) = 4^x_________________When Wombats Strike!
JMickle
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Posted: Tue Mar 02, 2010 5:10 pm    Post subject:

 psy_wombats wrote: Ha, whoops, I accidentally changed 2 things... That was supposed to be 2/x instead of 2x, but in any case, pretty sure the only solution to the printed equation was x=0. ln(4) yields 16 for the left hand side and something like 7 for the right. So we'll go back to what I intended:
take a natural log of both sides and you end up with 2x = x*ln(4)

divide by x and you get: x=ln(4)
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psy_wombats
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 Posted: Tue Mar 02, 2010 5:42 pm    Post subject: ln(4^x) = 2x*ln(x), not x*ln(4) That yields 2x = 2x * ln(4) Which means x must be zero._________________When Wombats Strike!
JMickle
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Posted: Tue Mar 02, 2010 5:54 pm    Post subject:

 psy_wombats wrote: ln(4^x) = 2x*ln(x), not x*ln(4)
What? How does it? When you take a log of something, any powers can be turned into multipliers outside the function, so the x becomes a multiplier of the log. I have no idea where you are getting the 2x from.
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psy_wombats
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 Posted: Tue Mar 02, 2010 6:00 pm    Post subject: ln(4) = 2 * ln(2) I was just simplifying so the division worked out, that wasn't really where the error was. You'd have come up with the same answer. You had: 2x = x*ln(4) Dividing by 2 yields: 2 = ln(4) not x = ln(4)_________________When Wombats Strike!
JMickle
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 Posted: Tue Mar 02, 2010 6:05 pm    Post subject: aw shit i'm an idiot_________________Hello I am a regular person!
quack_tape
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Posted: Tue Mar 02, 2010 6:12 pm    Post subject:

 psy_wombats wrote: ln(4^x) = 2x*ln(x), not x*ln(4)

Can you explain this bit one more time? I get why JMickle's wrong, but this...
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JMickle
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Posted: Tue Mar 02, 2010 6:33 pm    Post subject:

 quack_tape wrote: I get why JMickle's wrong, but this...
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psy_wombats
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Posted: Tue Mar 02, 2010 7:57 pm    Post subject:

quack_tape wrote:
 psy_wombats wrote: ln(4^x) = 2x*ln(x), not x*ln(4)

Can you explain this bit one more time? I get why JMickle's wrong, but this...

ln(4^x) = x*ln(4) = 2x*ln(2)

It just made the point clearer; ln(4^x) = x*ln(4) isn't "wrong" per se.
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Spock_Logic147
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 Posted: Tue Mar 02, 2010 8:04 pm    Post subject: new equation guys?_________________Here's the thing...
quack_tape
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Posted: Tue Mar 02, 2010 8:44 pm    Post subject:

psy_wombats wrote:
quack_tape wrote:
 psy_wombats wrote: ln(4^x) = 2x*ln(x), not x*ln(4)

Can you explain this bit one more time? I get why JMickle's wrong, but this...

ln(4^x) = x*ln(4) = 2x*ln(2)

It just made the point clearer; ln(4^x) = x*ln(4) isn't "wrong" per se.

It was the 2x*ln(x) that was screwing with my mind. I guess it was just a typo.
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Million Dollar Horse

"This is rather as if you imagine a puddle waking up one morning and thinking, 'This is an interesting world I find myself in - an interesting hole I find myself in - fits me rather neatly, doesn't it? In fact it fits me staggeringly well, must have been made to have me in it!'"
quack_tape
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 Posted: Tue Mar 02, 2010 8:48 pm    Post subject: Fine, new equation(s): x=2+y y=2x_________________Million Dollar Horse "This is rather as if you imagine a puddle waking up one morning and thinking, 'This is an interesting world I find myself in - an interesting hole I find myself in - fits me rather neatly, doesn't it? In fact it fits me staggeringly well, must have been made to have me in it!'" -Douglas Adams
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