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#1 |
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FFR Player
Join Date: Jul 2007
Location: toronto
Posts: 38
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Hello. I'm having difficulties solving this particular Exponential Equation and according to my understanding it involves logarithms.
3^2x - 6(3^x) + 9 = 0 I'm not interested in the answer. I already know that the answer is one but I am having difficulties manipulating this equation. I would really appreciate it if someone could show me what I should do because I have lots of homework with similar questions. Below I have provided some work I attempted on this question. 3^2x - 6(3^x) + 9 log 3^2x - [log 6 + log 3^x] = - log 9 2x log 3 - log 3^x = log 6 - log 9 x[log 3^2 - log 3] = log 6/9 Eventually, If i continue I end up with the wrong answer. |
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#2 |
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Very Grave Indeed
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I edited your thread title to follow the explicit thread titling requirements in the rules. That's your one warning to remember to follow directions.
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#3 | |
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Lombax Connoisseur
Join Date: May 2006
Location: Virginia
Age: 31
Posts: 2,548
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Quote:
You may have brought your x down wrong? Instead of : log 3^2x - [log 6 + log 3^x] = - log 9 I see it as : 2x log 3 - 6x log 3 = - log 9 -4x log 3 = - log 9 x = - log 9 / - 4 log 3 I am not quite sure, because we are doing exponential equations as well, but they are quite different from this. |
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#4 |
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Hunger Games Hunty
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Wait I'm sorry, nevermind. -.- Scratch everything I just said..
Last edited by Jtehanonymous; 01-19-2008 at 11:15 PM.. |
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#5 |
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FFR Player
Join Date: Jul 2007
Location: toronto
Posts: 38
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Thanks a lot buddy I got it. Let me post my work in a second for you to look over
2x log 3 - log 3^6x = -log 9 2x log 3 - 6xlog 3 = - log 9 2x (log 3 - 3log 3) = -log 9 2x (log 3/9) = - log 9 2x = - log 9 / log 1/3 2x = 2 x = 1 =) Thanks for the help flipsta_lombax and Jtehanonymous. Last edited by og4lif; 01-19-2008 at 11:16 PM.. |
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#6 |
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Lombax Connoisseur
Join Date: May 2006
Location: Virginia
Age: 31
Posts: 2,548
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I don't agree with you jtehanonymous. I just don't see where you need to take the log of 6 when all you need to take the log of is that 3^x.
Oh btw, my answer: x = - log 9 / - 4 log 3 x = log 9 / 4 log 3 x = .5 The logistics seem quite reasonable on my behalf. |
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#7 |
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Hunger Games Hunty
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I realized that when I looked over my post again, and that's why I felt stupid and edited it. XD
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#8 |
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FFR Player
Join Date: Jul 2007
Location: toronto
Posts: 38
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.5 doesn't seem to work if i place it back into the equation.
But yeah thanks again guys. So does this thread get locked up now? |
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#9 | |
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Hunger Games Hunty
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Quote:
I dunno, I thought the 3log3 would make a log 27 Edit : 2x (Log 1/9) = -log9 2x = -log9/log1/9 (Which equals 1) 2x = 1 x = 1/2 So I agree with flip. XD Last edited by Jtehanonymous; 01-19-2008 at 11:25 PM.. |
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#10 |
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FFR Player
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...What math level is this? Algebra II? I'm in Accelerated Geometry, I can't figure it out, x.x
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#11 |
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Very Grave Indeed
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Yay the successful conclusion of our inaugural thread.
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#12 | |
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FFR Player
Join Date: Aug 2006
Location: Storm Sanctuary!
Posts: 255
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Quote:
If you look carefully, this question is somewhat different from your average exponential equation question in that it is similar to a quadratic equation. Picture this: instead of 3^x, think of it as x and 3^2x as X^2. Everything in this equation should look like (X^2)-6(x)+9=0 If you solve the quadratic properly, you should factor and get (x-3)(x-3)=0 Now think of it in terms of the actual equation. You should get factors of ((3^x)-3)((3^x)-3)=0 From here, you should solve for (3^x)-3=0 in the usual fashion by the following: 1. adding 3 to both sides to make this: (3^x)=3 2. do log base 3 on both sides to look like this: Log3(3^x)=Log3(3) 3: Your x value should be equal to Log3(3) or 1 Always check your answer by plugging it into the original equation. Last edited by Master_of_the_Faster; 03-13-2008 at 04:12 PM.. |
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#13 |
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(+ (- (/ (* 1 2) 3) 4) 5)
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Master of the Faster: Conclusion usually means no bumping 2 months after the fact.
Edit: I suppose, though, that adding to the discussion doesn't really do much harm, especially in such a slow moving subforum. Whatever. Last edited by sumzup; 03-18-2008 at 09:55 PM.. Reason: clarification & rethinking |
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