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Find Domain Of: √(log1/5(1/Sin5x))

  1. Apr 8, 2012 #1
    1. The problem statement, all variables and given/known data
    √(log1/5(1/Sin5x))

    Log to the base 1/5*(1/Sin5x)


    FIND THE DOMAIN


    I have an answer to this solution, I Just need some help with understanding it :(... and also could someone tell me if info is missing.


    f(x) = √(log1/5(1/Sin5x))

    Restrictions:

    1/sin5x ǂ 0

    sin5x ǂ 0

    5x = ∏n
    xǂ (∏n)/5

    √(log1/5(1/Sin5x)) ≥ 0
    (square both side)
    (log1/5(1/Sin5x)) ≥ 0

    1/sin5x ≤ 1

    Sin5x ≥ 1

    **** Sin M ≤ 1 ***** (this is refferring to the slope)


    Sin5x = Sin(∏/2 + 2∏n)

    5x = = ∏/2 + 2∏n
    x = ∏/10 + 2∏n/5

    ans: XE(∏/10 + 2∏n/5]

    anyways i would appreciate if some one could help me understand this solution better. <3
     
  2. jcsd
  3. Apr 8, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I assume you mean log base 1/5:
    [tex]\sqrt{log_{1/5}(\frac{1}{sin(5x)})}[/tex]
    Yes, the sin(5x) in the denominator cannot be 0. Sine sin(x) is 0 for x any multiple of [itex]\pi[/itex], that means 5x cannot be as multiple of [itex]\pi[/itex]:[itex]x\ne n\pi/5[/itex] for any integer n, as you say.

    But also, logarithm is only defined for positive numbers so in addition to [itex]sin(5)\ne 0[/itex], we must have sin(5x)> 0. [itex]sin(\theta)[/itex] is positive for [itex]0< x\le \pi[/itex], [itex]2\pi< x< 3\pi[/itex], etc. We can write that as [itex]2n\pi< x< (2n1)\pi[/itex] for any integer n.

    I don't know why you are looking at "[itex]\sqrt{log_{1/5}(1/sin(5x)}\ge 0[/itex]". That is a statement about the value or range of the function and has nothing to do with the domain. Rather, what is important is that in order that a square root exist, the argument must be non-negative. And [itex]\log(x)\ge 0[/itex] if and only if [itex]x\ge 1[/itex]. So we must have [itex]log_{1/5}(1/sin(5x))\ge 1[/itex] which means that we must have [itex]1/sin(5x)\ge 1/5[/tex] which, since we have already required that it be positive, is the same as [itex]5\ge sin(5x)[/itex]. For what x is that true?
     
  4. Apr 9, 2012 #3
    Yea my bad D:
    This is clear.
    When
    5≥sin5x

    For all X values, Sin5x will be less than 5..
     
    Last edited: Apr 9, 2012
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