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How do i solve: logx^2 = (logx)^2?

  1. Oct 21, 2007 #1
    i do not know how to do this at ALL!!
     
  2. jcsd
  3. Oct 21, 2007 #2
    2 * log (x) = log(x) ^ 2
    Log(x) = 0 is root
    OR
    2 = log(x)
    Then
    x=1
    OR
    x=100
     
  4. Oct 21, 2007 #3
    i dont get it....
     
  5. Oct 21, 2007 #4
    how about this:

    [tex]\ln (x^2) = 2 \ln (x) = \ln (x) \ln (x)[/tex]
    [tex]2 = \ln (x)[/tex]
    [tex]x = e^2[/tex]
    and of course x=1
     
  6. Oct 21, 2007 #5
    so is it like:

    logx^2 = (logx)^2
    2logx = (logx)(logx)
    2 = logx
    x = 10^2??? --> x = b^y = y = b^x ??? is that it??
     
  7. Oct 21, 2007 #6
    how about this: 2^y + 5^y-2??

    do i subtract??
    like:
    2^y = -5^y-2 ???
     
  8. Oct 22, 2007 #7

    Gib Z

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    Homework Helper

    You could think of it as a quadratic equation in log x.
     
  9. Oct 22, 2007 #8

    HallsofIvy

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    That's one difficulty with not showing any work! Is "log x" the common logarithm or the natural logarithm? In "elementary" work, it is standard to use "log" to mean the common logarithm (base 10) and "ln" to mean the natural logarithm (base e). In more advanced work, it is standard to use "log" to mean the natural logarithm and not use the common logarithm at all.

    Rogerio was assuming your "log" was the common logarithm. Since your question made it look like you didn't know any thing about logarithms, he assumed it was "elementary" mathematics. Your answer, then used "log" to mean natural logarithms. The only way WE can know which is correct is for you to tell us and the only way for YOU to know is to check your text book.
     
  10. Oct 22, 2007 #9

    Gib Z

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    Unless of course the answer is intended to be left in the log form =]
     
  11. Jan 26, 2009 #10
    this question is wrong
    logx^2=2logx=logx+logx
    and
    (logx)^2= logx*logx
    and
    logx*logx =/= logx + log x

    as simple as
    x*y =/= x+y
     
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