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What is the domain of f(x)=ln(ln(ln(x)))?

  1. Sep 13, 2007 #1
    Can anyone explain how to get the domain of f(x)=ln(ln(ln(x)))?

    I am stuck at e^lnlnx > e^o...

    Thank you.
  2. jcsd
  3. Sep 13, 2007 #2


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    simplify the left side... and the right side is just 1.
  4. Sep 13, 2007 #3


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    Start with the outer function!

    We must have ln(ln(x))>0
    Hence, we must have

    whereby you should readily find the domain of x!
  5. Sep 13, 2007 #4
    Can someone explain the steps, I REALLY don't get this and it has been bothering me for a week.
  6. Sep 13, 2007 #5
    As arildno said, determine what values of the inner functions belong to the domains of the outer functions: [tex]
    \ln \ln \ln x \in \mathbb{R} \Rightarrow \ln \ln x > 0 \Rightarrow \ln x > 1 \Rightarrow x > e[/tex]
  7. Sep 13, 2007 #6


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    Ok, as this has been going on for two threads now. For ln(ln(ln(x))) to be defined, the argument of the leftmost ln must be > 0. So ln(ln(x))>0. Exponentiate both sides. exp(ln(ln(x)))=ln(x), exp(0)=1. Hence ln(x)>1 since exp(x) is monotone. Exponentiate again. exp(ln(x))=x. exp(1)=e. So x>e. Since exp(x) is monotone still. This is exactly the same thing everyone else has been saying. What don't you 'get'?
  8. Sep 14, 2007 #7


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    e^(lnlnx) = lnx (because e^(lnr) = r for any r)

    so substituting lnx instead of e^lnlnx into your inequality:


    lnx>1 (since e^0 = 1)


    e^(lnx) > e^1 (take e to the power of both sides)

    we know that e^(lnx) = x (same reason as before)

    so substitute that into the previous inequality...

    x > e^1

    x > e
  9. Sep 14, 2007 #8
    Thank you physics!!!!!!!!!!1
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