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Math Help - Constant plus natural log of constant. Find constant?

  1. #1
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    Constant plus natural log of constant. Find constant?

    On the title, I meant variable.

    I need to know how to solve for a variable when you have a variable plus a number multiplied by the natural log of the same variable, and it equals some incredibly complex number. It's constant, as I stated, so that side is pointless and can be treated as c.

    so... how do I solve for x in a situation such as...

    x^2+2ln(x)=SomeNumber

    Make somenumber anything you want. Mine specifically is is between 8 and 9. It is just the quantity of a number plus a number multiplied by the natural log of another number end quantity all over 6. Ignoring that, could anyone help?

    This is actually something simple I think I should know. I am just trying to wrap up a much more complicated calculus problem. I just forgot how to do this and cannot seem to find it anywhere.
    Last edited by purdyaw; February 29th 2008 at 12:49 PM.
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  2. #2
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    x^{2}+2ln(x)=a is kind of difficult to solve for x algebraically.

    You can always try Newton's method to solve for a particular value of x.

    Suppose you had x^{2}+2ln(x)=3

    Try an initial guess of 1.5

    1.5-\frac{(1.5)^{2}+2ln(1.5)-3}{2(1.5)+\frac{2}{1.5}}=1.485939...

    Keep iterating unitl you hone in. This converges pretty fast, so not many are needed to get it within a few decimal places.
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  3. #3
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    Have you ever heard of the LambertW function?. That can be helpful when it's difficult to solve algebraically.

    For instance, to solve your y=x^{2}+2ln(x) for x, we get:

    x=e^{\frac{-1}{2}LambertW(e^{y})+\frac{y}{2}}
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  4. #4
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    Thanks. I am going to go with the first way you explained it, though.
    Last edited by purdyaw; February 29th 2008 at 01:55 PM.
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  5. #5
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    I have not had a lot of experience with it either. But, I can show you a basic example. I will make one up.

    Something like 4^{t}=9t. Solve for t.

    The idea is to get it into the form A=Be^{B}

    4^{t}=9t

    1=\frac{9t}{4^{t}}

    1=9te^{-tln(4)}

    \frac{1}{9}=e^{-tln(4)}

    \frac{-ln(4)}{9}=(-tln(4))e^{-tln(4)}....required form A=Be^B

    -tln(4)=W(\frac{-ln(4)}{9})

    t=\frac{W(\frac{-ln(4)}{9})}{ln(4)}

    As far as getting a numerical value for this, I think there is a table you look them up in.

    This case works out to be t=2.130588... or t=.1337459....
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  6. #6
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    What I have is

    <br />
c^2+3ln(c)=\frac{20+27ln(3)}{6}<br />

    I am on the average value of a function, which is quite easy... The right side is the average value with certain bounds.
    The left side is the function as translated into c, where c = a place where the functions hits it. That function being:

    <br />
f(x)=x^2+3ln(x)<br />
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