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 08-03-08, 05:03 PM #1 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) Logarithm To solve for x I thought I would rewrite the equation in exponential form: 3^(2-x) = 3 Anything raised to the power of 1 is itself. So, x must equal 1. Because 3^(2-1) = 3 Why is this incorrect? __________________ C://dos C://dos.run run.dos.run
08-03-08, 05:07 PM   #2
crash66
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 Originally Posted by phantomcow2 To solve for x I thought I would rewrite the equation in exponential form: 3^(2-x) = 3 Anything raised to the power of 1 is itself. So, x must equal 1. Because 3^(2-1) = 3 Why is this incorrect?

You're lucky so many people here like to show off their mad math skillz, otherwise you'd have spent a fortune on tutoring by now.

 08-03-08, 05:15 PM #3 apricissimus  L T X B O M P F A N S R     Join Date: Jun 2008 Location: Malden, MA Bikes: Bianchi Volpe, Bianchi San Jose, Redline 925 Posts: 2,334 Mentioned: 2 Post(s) Tagged: 0 Thread(s) Quoted: 1638 Post(s) log_3(2 - x) = 3 3^log_3(2 - x) = 3^3 2 - x = 9 x = -7 Last edited by apricissimus; 08-03-08 at 05:19 PM.
 08-03-08, 05:41 PM #4 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) I actually solved that a short bit ago. Thanks though. __________________ C://dos C://dos.run run.dos.run
 08-03-08, 05:55 PM #5 deraltekluge Senior Member     Join Date: Sep 2006 Bikes: Kona Cinder Cone, Sun EZ-3 AX Posts: 1,195 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Uh...3³ is not 9. It's 27.
08-03-08, 05:56 PM   #6
phantomcow2
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 Originally Posted by deraltekluge Uh...3³ is not 9. It's 27.
I now realize that. THanks
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 08-03-08, 06:02 PM #7 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) Okay okay, now how about this: The best I can do is rewrite it as .5xLog(.7) = 5log(1-x) __________________ C://dos C://dos.run run.dos.run
 08-03-08, 06:22 PM #8 JaRow trois, mon frère     Join Date: Jan 2008 Location: Gainesville/Miami, FL Bikes: '01 Gary Fisher Wahoo, '08 Giant TCR C2 Posts: 576 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Is this your math hw?
 08-03-08, 06:28 PM #9 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) Yes, and let me write my disclaimer now: I seek guidance only, not the answer. __________________ C://dos C://dos.run run.dos.run
 08-03-08, 06:31 PM #10 Taerom Hazardous     Join Date: Jun 2005 Location: Quarantine Bikes: 2005 Trek Liquid 55, 2009 Haro Mary SS Posts: 727 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Just plug it into your calculator's numeric solver and be done with it.
 08-03-08, 06:31 PM #11 deraltekluge Senior Member     Join Date: Sep 2006 Bikes: Kona Cinder Cone, Sun EZ-3 AX Posts: 1,195 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) That's not correct. You did the left side of the equation correctly, but you got things backwards on the right side. Remember, log(x^y) = y* log(x)
08-03-08, 06:33 PM   #12
phantomcow2
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 Originally Posted by Taerom Just plug it into your calculator's numeric solver and be done with it.
That's what made me mathematically handicapped in the first place!
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08-03-08, 06:36 PM   #13
apricissimus
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 Originally Posted by deraltekluge Uh...3³ is not 9. It's 27.
*slaps palm on forehead*

I'm actually a math editor in real life. Minus a million points for me.

Last edited by apricissimus; 08-03-08 at 06:40 PM.

 08-03-08, 06:39 PM #14 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) Oh, so I should have .5xlog(.7) = 1-xlog(5) then divide by log5 on both sides to begin isolation of the variable? __________________ C://dos C://dos.run run.dos.run
08-03-08, 06:45 PM   #15
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 Originally Posted by phantomcow2 Oh, so I should have .5xlog(.7) = 1-xlog(5) then divide by log5 on both sides to begin isolation of the variable?
Remember that it's (1-x). Make sure you keep your parentheses arranged properly.

08-03-08, 06:56 PM   #16
JaRow
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 Originally Posted by phantomcow2 Yes, and let me write my disclaimer now: I seek guidance only, not the answer.
I'm not faulting you for it. It's a shame more people aren't this proactive about their education.

Last edited by JaRow; 08-03-08 at 07:07 PM. Reason: bad spelling mistake

08-03-08, 07:04 PM   #17
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 Originally Posted by phantomcow2 Okay okay, now how about this: The best I can do is rewrite it as .5xLog(.7) = 5log(1-x)
By some algebraic manipulation, you can rewrite the equation as (35/2)^x = 5^2. (Try to do it yourself.)

From here take the log (to a certain base b) of both sides, and use the change of base formula, log_b(x) = log_a(x)/log_a(b) (or ln(x)/ln(b) if you like natural logarithms. And who doesn't?).

08-03-08, 07:17 PM   #18
phantomcow2
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 Originally Posted by apricissimus By some algebraic manipulation, you can rewrite the equation as (35/2)^x = 5^2. (Try to do it yourself.) From here take the log (to a certain base b) of both sides, and use the change of base formula, log_b(x) = log_a(x)/log_a(b) (or ln(x)/ln(b) if you like natural logarithms. And who doesn't?).
And as usual, it's the algebraic manipulation which sinks me.
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08-03-08, 07:23 PM   #19
apricissimus
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 Originally Posted by phantomcow2 And as usual, it's the algebraic manipulation which sinks me.
Here's the first couple of steps:

Rewrite as

(0.7^x)^1/2 = 5*(5^(-x))

Then square both sides...

0.7^x = (5^2)*(5^(-2x))

etc.