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Log inside log -- find the x in its max domains?

  1. Jul 1, 2017 #1
    1. The problem statement, all variables and given/known data

    15_Mat_B_1.4.png
    2. Relevant equations
    log_2 x = y
    2^y = x
    3^2^y

    3. The attempt at a solution
    log_2 x = y
    2^y = x
    log_2 {log _3 {log _2 { log_3 {2^y} } } }
    what am I suppose to do?
     
  2. jcsd
  3. Jul 1, 2017 #2

    fresh_42

    Staff: Mentor

    Try to work from outside in.

    ##f(a)= \log_2(a)## for some ##a##. Where is it defined, i.e. which values for ##a## are allowed? Say allowed the values are the set ##A_a##.

    Next you have ##f(b) = \log_3(b)## where ##f(b) \in A_a##. What does this mean for ##b##? Say we get allowed values in ##A_b##.

    Next you have ##f(c) = \log_2(c)## where ##f(c) \in A_b##. What does this mean for ##c##?

    And so on, until ##f(x)=\log_2(\log_3(\log_2(\log_3(\log_2(x))))) = f(e)##. The set ##A_e## is the solution.
     
  4. Jul 2, 2017 #3
    all I know the ' a ' must be greater than zero.
    you meant to work it inside out? because the f(e) = is the outside equation , and f(a) = the first log equation.
     
  5. Jul 2, 2017 #4

    fresh_42

    Staff: Mentor

    I meant what I wrote: outside in. (Maybe I had a step too many or a set ##A_*##.)
    We have ##f(x) = \log_2(sth.)## at the start. I called that something ##a## which actually is ##a=\log_3(\log_2(\log_3(\log_2(x))))##, but I don't care by now. Then we have, as you've said, ##a > 0##. Now ##a = \log_3(b) > 0## for a new something called ##b##, which is ##b=\log_2(\log_3(\log_2(x)))##. Which ##b## are here allowed? Etc.
     
  6. Jul 2, 2017 #5
    ##f(x) = \log_2(a)##
    ##a=\log_3(\log_2(\log_3(\log_2(x))))##,
    a>0
    ##a=\log_3(\log_2(\log_3(\log_2(x))))>0##,
    ##a=\log_3(b)>0##,
    3^a = b

    ##b=\log_2(\log_3(\log_2(x)))>0##,
    ##b=\log_2(c)>0##,
    2^b = c --> 2^(3^a) = c
    c>0
    ##\log_3(\log_2(x)))>0##

    ##c = \log_3(\log_2(x)))>0##
    c = ##\log_3(d)>0##
    3^c = d
    3^2^(3^a) = d
    d>0
    ##(\log_2(x)))>0##

    ##d=(\log_2(x)))##
    2^d = x
    2^3^2^(3^a) = x
    x>0

    ???
     
  7. Jul 2, 2017 #6

    fresh_42

    Staff: Mentor

    If ##a = \log_3(b) > 0##, then ##b=3^a>3^0=1##. And if ##b = \log_2(c) > 1## then ##c>2## ...
     
  8. Jul 2, 2017 #7
    @fresh_42 thanks! I never thought i'll solved it
     
  9. Jul 2, 2017 #8
    why not include
    f(x) = log 2 (a)
    2^a = f(x) ?
     
  10. Jul 2, 2017 #9

    fresh_42

    Staff: Mentor

    You included it, as you said ##a>0##, which is the first condition. Here ##f(a)=\log_2(a)## has still all reals as possible values. But in order to have a number ##a>0## written as another logarithm, the next one ##a=\log_3(b)\; , \;b>0## isn't sufficient, because e.g. ##\log_3(\frac{1}{3}) = -1## which isn't positive. So ##b## has to be at least ##1##.
     
  11. Jul 3, 2017 #10
    I meant something like this :
    your answer : f(x) = ##log_2 (log_3 (log_2 (log_3(log_2 ( x ) ) ) ) ) ##
    ##f(x) = log_2 (a) ; a > 0##
    ##a = log_3 (log_2 (log_3(log_2 ( x ) ) ) )##
    ##a = log_3 (b)##
    b = 3^a
    for a > 0, b > 3^0 ; b > 1

    ##b = log_2 (log_3(log_2 ( x ) ) )##
    ##b = log_2 (c)##
    c = 2^b
    ##b>1 ; c > 2##

    ##c = log_3(log_2 ( x ) ) ##
    ##c = log_3 (d)##
    ##d = 3^c##
    ##c>2; d>3^2; d>9 ##

    ##d = log_2 ( x ) ##
    ##x = 2^d##
    ##d>9; x > 2^9 ; x > 512
    ##

    why not include f(x) ?
    ##f(x) = log_2 (a) ; ##
    f(x) > 0
    a = 2^(fx) ; a > 1
    and b > 3
    c > 8
    ##d > 3^8 ##
    x>2^##3^8##

    oh, is it the f(x) not need to be more than zero, because after f(x) there's no logarithm anymore. so f(x) for any real numbers, while a need to be ' more than zero ' condition because of logarithm equation?
     
  12. Jul 3, 2017 #11

    fresh_42

    Staff: Mentor

    Not sure I understand your question, but ##512## is the correct answer. The point is that ##f(x)## is written as a logarithm. This means we need a positive argument for this logarithm. Now this logarithm is again written as a logarithm, and it doesn't only need a positive argument, it needs a positive function value, too. Thus it has to be greater than one. Now this logarithm is again written as a logarithm, which means it needs a positive argument and a function value greater than one, which means its argument has to be greater than the base. And so on.

    The stronger (as just being positive) requirements all come from the fact, that in order for a logarithm to have positive values, it has to be greater than one. In order for a logarithm to have values greater than one, it has to be greater than the base ##b##. In order for a logarithm to have values greater than ##b##, it has to be greater than ...
     
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