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How do i solve the following with or without use of logs?

  1. Oct 3, 2009 #1
    How do i solve the following with or without use of logs?
    3^x=x^2

    the problem being that i have x both in my exp and in my base, what i have done is taken each side and gotten

    (3^x)^(1/x)=(x^2)^(1/x)
    3=x^(2/x)
    3^(1/2)=(x^(2/x))^(1/2)
    sqrt(3)=x^(1/x)

    0.5*ln(3)=(1/x)*ln(x)
     
  2. jcsd
  3. Oct 3, 2009 #2

    LCKurtz

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    You can't solve it algebraically. It can be solved in terms of a non-elementary function called the Lambert W function. Numerically the answer is -.6860267245.
     
  4. Oct 3, 2009 #3
    on my calculator i can get x=1.0434693552 but cannot prove it, is there no other way other than Lambert W function?
     
  5. Oct 3, 2009 #4

    LCKurtz

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    No. That's it. Numerically, which is how the Lambert W function's values are gotten anyway. And your x=1.0434693552 doesn't solve 3x = x2.
     
  6. Oct 3, 2009 #5
    no sorry my mistake, that was for another excercise,
    i looked up lambert W functions and i got this


    3x=x2
    (31/2)x=x
    1=x/(31/2)x =====> (31/2)=a
    1=x/ax
    1=x*e-x*ln(a)
    -ln(a)=[(-ln(a)*x)*e(-ln(a)*x)

    after applying W to each side

    -ln(a)*x=W(-ln(a))
    x=-W(-ln(a))/(ln(a))

    is this correct, and also how do i get a numerical answer for this??
     
    Last edited: Oct 3, 2009
  7. Oct 3, 2009 #6
    i found an online calculator that gave me the value, which looks right, but it only gives me one value and i believe there are 2. how can i work it out?
     
  8. Oct 3, 2009 #7

    Mentallic

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    No, there is only 1 real value.
     
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