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Math Help - Amc 12b #20

  1. #1
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    Amc 12b #20

    Let x be chosen at random from the interval (0,1). What is the probability that [\log_{10}(4x)]-[\log_{10}(x)]=0?

    Here [x] denotes the greatest integer that is less than or equal to x.

    This is multiple choice, but I don't think posting the possibilities are necessary.
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    Quote Originally Posted by Jameson
    Let x be chosen at random from the interval (0,1). What is the probability that [\log_{10}(4x)]-[\log_{10}(x)]=0?

    Here [x] denotes the greatest integer that is less than or equal to x.

    This is multiple choice, but I don't think posting the possibilities are necessary.
    I do not have much time now but I was thinking maybe you can do this.
    [x]=[x+k],0<k<1
    Then, it seems to me (I did not formally prove it) then,
    j<x<1-k+j
    are all solutions for each integer j.

    Thus,
    [\log (4x)]-[\log x]=0
    Thus, since 0<\log 4<1
    [\log 4+\log x]=[\log x]
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  3. #3
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    Let, us solve, for 0<x<1
    [\log 4x]=[\log x]
    We have,
    [\log 4+\log x]=[\log x]
    Thus, all solutions for u=\log x are,
    u\in \bigcup_{j\in\mathbb{Z}}(j,1-\log 4+j]
    Thus, we that, all solutions satisfies
    j<\log x\leq 1-\log 4+j
    Iff,
    10^j<x\leq 2.5\cdot 10^j
    .
    Note, j cannot be zero or positive because it would violate the inequality 0<x<1.
    Thus, j=-1,-2,-3,-4,... all work.
    Thus,
    .1<x\leq .25 thus, length =.15
    .01<x\leq .025 length=.015
    .001<x\leq .0025 length=.0015
    and so on "ad infinitum" (I am so cool using latin phrases).
    Thus, we have the total length of the solutions to be:
    .15+.015+.0015+....=.16666=1/6
    this is a regular infinite geometric series.

    My point is that, we can "intuitively" think of probability as the length of the success (which are the solutions) divided by the total possibilities (which is the length of interval). Thus, the probability is \frac{1}{6}.
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