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Logarithm equations home

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data

    i.2(lgx)^2 - lgx = 0

    and

    ii. lg(2x-2)^2= 4lg(1-x)

    and

    iii. lgx-6 / lgx-4 = lgx.

    I simply do not manage to solve these equations, and I would therefore be happy for all help. Thanks in advance.
     
  2. jcsd
  3. Oct 17, 2012 #2
    Hints:

    i) and iii) Make the substitution z = lg x, and solve for z. What do you get?

    ii) Notice that: lg (2x - 2) = lg[(-2)(1 - x)]. Now, if lg represents a principal value of the logarithm with a branch cut along the negative real axis, then:

    [tex]
    \mathrm{lg} \left[(-2)(1 - x) \right] = \mathrm{lg}(e) \cdot \ln{\left[(-2)(1-x) \right]} = \left\lbrace
    \begin{array}{ll}
    \mathrm{lg}(e) \cdot [\ln(2) + \ln(1-x)] = \mathrm{lg}(2) + \mathrm{lg}(1 - x) &, \ \mathrm{Arg}(1-x) \le 0 \\

    \mathrm{lg}(e) \cdot [\ln(2) + \ln(1-x) - 2 \pi \, i] = \mathrm{lg}(2) + \mathrm{lg}(1 - x) - 2 \pi \, \mathrm{lg}(e) \, i &, \ \mathrm{Arg}(1-x) > 0
    \end{array}\right.
    [/tex]

    Then, substitute z = lg(1 - x) and solve the equsation for z. Make sure that the condition for Arg(1 - x) is satisfied in the proper case.
     
  4. Oct 17, 2012 #3
    This is the only one I can't figure out. Isn't there an easier way?
     
  5. Oct 17, 2012 #4
    First of all, what do you mean by "lg"? Second, what is the domain in which you solve this equations? Is it complex numbers, or are you restricted to real numbers only?
     
  6. Oct 17, 2012 #5

    Lg = Logarithm
    Real numbers only, but no worries, I think I worked it out.

    Thanks!
     
  7. Oct 17, 2012 #6

    Ray Vickson

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    (ii) Is ambiguous. Do you mean
    [lg(2x-2)]^2 = 4 lg(1-x), or do you mean
    lg[(x-2)^2] = 4 lg(1-x)?

    (iii) As written, (iii) is
    (lg x) - (6/lg x) - 4 = lg x. Did you mean that? Or, did you mean
    lg(x-1)/lg(x-4) = lg x? Or did you mean
    [lg(x-1)/lg x] - 4 = lg x?

    You need to use brackets to make things clear.

    RGV
     
  8. Oct 17, 2012 #7
    My mistake. Check the link. http//bildr.no/view/1297631
     
    Last edited: Oct 17, 2012
  9. Oct 17, 2012 #8

    Ray Vickson

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    Science Advisor
    Homework Helper

     
    Last edited by a moderator: May 6, 2017
  10. Oct 17, 2012 #9
    I don't know actually, but I think it is logarithms to base 10.

    And regarding you question, please check out the link attached above.

    The correct answers are:

    iii. X = 1000 or X = 100

    ii. x =-1

    However, it is not that important since I've got until monday.
     
  11. Oct 17, 2012 #10
    If the symbol is given as logx it is the default one, with the base of 10. In other cases you have the logarithm given as lognX - where n is the base and it is equal to log10X/log10n and finally you have the natural logarithm ln X which has a base of e - so ln X = logeX. Hope that clarifies it for you :)
     
  12. Oct 17, 2012 #11

    Ray Vickson

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    Science Advisor
    Homework Helper

    Unfortunately, what you say is not completely correct: many authors use log x to mean ln x, etc., so you really do need to check. Often if one wants to use a base other than e, one uses a subscript like log10(x) or log2(x) or an index notation like log[10](x), etc. I think usage has changed a lot during the more than 50 years since I was in school; then log(x) really did always mean log[10](x). Now, not so much.

    RGV
     
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