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anonymous

  • one year ago

Anyone knows how to solve this??

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  1. anonymous
    • one year ago
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    \[\frac{ (x + 5) }{ 3 } = \frac{ \ln(x) }{ -2 }\]

  2. anonymous
    • one year ago
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    I'm really confused how to go on about this. I tried doing it like this: \[-2(x + 5) = 3\ln(x)\] \[-2x -10 = \ln(x^3)\] \[e ^{-2x -10} = x^3\] and after I'm just stuck.. ._.

  3. anonymous
    • one year ago
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    i forget how to do this sry

  4. mathmate
    • one year ago
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    hint: Examine what you just wrote: −2x−10=ln(x3)

  5. mathmate
    • one year ago
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    actually : \(−2x−10=ln(x^3)\)

  6. mathmate
    • one year ago
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    hint: graph (by hand) each side of the equation \(−2x−10=ln(x^3)\)

  7. anonymous
    • one year ago
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    well it does imply that e has to have an exponent of -2x-10 to give x^3 so \[e ^{-2x}*\frac{ 1 }{ e ^{10} } = \ln(x^3)\]

  8. mathmate
    • one year ago
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    |dw:1440154856949:dw| Do they ever intersect?

  9. anonymous
    • one year ago
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    \[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]Choose a suitable approximation.

  10. anonymous
    • one year ago
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    @mathmate yes they do intersect which means that they have one 1 root but I want to find the value for x in order to find the coordinate of the intersection.

  11. anonymous
    • one year ago
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    @Audio I'm not sure what that is, can you please expand on that?

  12. anonymous
    • one year ago
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    It is, indeed, an expansion. :) http://mathworld.wolfram.com/SeriesExpansion.html

  13. anonymous
    • one year ago
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    The best bet here is to graph though. You never know how inaccurate an answer can be when you use Taylor Expansions to solve transcendental equations.

  14. mathmate
    • one year ago
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    @nopen Are you working on calculus or numerical methods?

  15. anonymous
    • one year ago
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    um the working that you see in my first comment is what I used to approach this problem.

  16. mathmate
    • one year ago
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    This is basically a numerical methods problem, in which you know there is a root, and need a method to refine the value accurately. This is the approach.

  17. anonymous
    • one year ago
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    ah, yes that's exactly what I wanted to do actually. But I don't know really know how, as I'm stuck at the 3rd step

  18. mathmate
    • one year ago
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    The third step is the meat of the problem. start with something simple,like f(x)=-2(x+5)-3log(x) and calculate f(0.01) and f(0.05) to confirm that there is a root.

  19. anonymous
    • one year ago
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    I'm getting f(0.01) = -4.02 and for f(0.05) = -6.196

  20. anonymous
    • one year ago
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    but wait how are you taking random numbers for the domain like that?

  21. mathmate
    • one year ago
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    Sorry, I usually use log(x) to mean ln(x), and log10(x) when it's to base 10. We use log10 so rarely that I use log to mean log e.

  22. mathmate
    • one year ago
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    They are not random numbers, they are obtained from the graph that you said earlier that there is a root.

  23. mathmate
    • one year ago
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    From graphing, we know the value of x is very small, such that -2(x+5) is about -10.

  24. anonymous
    • one year ago
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    oh. Oops okay then f(0.01) = 3.79 and f(0.05) = -1.11

  25. mathmate
    • one year ago
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    Anyway, have you recalculated f(0.01) and f(0.05), which should have alternate signs.

  26. mathmate
    • one year ago
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    Good!

  27. anonymous
    • one year ago
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    yup

  28. mathmate
    • one year ago
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    Now, we can do one of the two things. 1. by trial and error, or bisection method: Keep calculating f(x) while narrowing down between 0.01 and 0.05 until you get f(x) as close to 0 as you wish. You'd better have a programmable calculator or a computer for that.

  29. mathmate
    • one year ago
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    For example, f(0.01)=3.795510557964274 f(0.05)=-1.112803179338027 f(0.035)=-0.01277834752183082 make guesses along the way f(0.03465)=0.01807266003867447 f(0.03475)=...

  30. mathmate
    • one year ago
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    are you still with me?

  31. anonymous
    • one year ago
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    Yes, but I see that's gonna be a long process by hand :S

  32. mathmate
    • one year ago
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    Good, now you will appreciate 2. by Newton's method.

  33. mathmate
    • one year ago
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    2. Newton's method says given f(x)=-2(x+5)-3log(x) and an initial approximation x0, under certain conditions of convergence (which we satisfy here), a better approximation x1=x0-f(x0)/f'(x0) where f'(x)=d f(x)/d x = -2 - 3/x in this case

  34. mathmate
    • one year ago
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    and we can repeat the process until \(x_n\) is identical to \(x_{n+1}\). The best part is that the number of correct decimal places doubles after each iteration because it has second order convergence.

  35. anonymous
    • one year ago
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    Oooh I see what you're doing now

  36. mathmate
    • one year ago
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    Do you want me to get you started, or that's good enough?

  37. anonymous
    • one year ago
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    Oh that's good enough thank you very much, because the original question just asked to mention if there is any roots if so state the number of them, so I was just curious to find the coordinate for it

  38. anonymous
    • one year ago
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    thanks! @mathmate

  39. mathmate
    • one year ago
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    If you are curious to use the method, you can check your answer : x=0.03485461440500114

  40. mathmate
    • one year ago
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    If you are curious to use the method, you can check your answer : x=0.03485461440500114 accurate to the last digit, after just a few iterations.

  41. anonymous
    • one year ago
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    lol that's not a few xD

  42. mathmate
    • one year ago
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    you're welcome! :) by the way, it took 4 iterations, starting from 0.03, the average of 0.01 and 0.05.

  43. anonymous
    • one year ago
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    oh oops I made a mistake with the starting value

  44. mathmate
    • one year ago
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    Exactly, the key is to find a good starting value! But any one reasonably close to the final root is ok, depending on the problem, and the conditions of convergence.

  45. anonymous
    • one year ago
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    ah now I got it right thnx!

  46. mathmate
    • one year ago
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    Good, have fun with math! :)

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