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Math Help - logarithm

  1. #1
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    logarithm

    could you possibly help me out? I am stuck on two questions
    the general term of an infinite sereis is tn= ( pie square) (3/11)^ n+1. how many terms must be added to reach a sum that is greater than 0.32?
    t1 = (pie square) (3/11)2 = 0.233672181
    r = (3/11)
    Sm = a * (1-rm)/(1-r)
    0.32 * 8/11 = 0.233672181 * (1-rm)
    I know I have to use logarithm but I get confused!

    my second question!

    convert 0.12 ( repeating decimal ) to a fraction using algebraic method!

    I know that I would use the fomula :
    s= a/ 1-r or am I totally wrong? can you guiys explain to me how I woudl do this question?
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  2. #2
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    Hello, Andrew!

    I know two ways to solve #2 . . .


    Convert 0.\overline{12} to a fraction using algebraic methods.

    Method #1: Geometric Series

    We have: . S\:=\:0.121212\hdots \:=\:0.12 + 0.0012 + 0.000012 + \hdots

    That is: . S \:=\:\frac{12}{10^2} + \frac{12}{10^2} + \frac{12}{10^6} + \hdots

    We have a geometric series with first term a = \frac{12}{100} and common ratio r = \frac{1}{100}

    Its sum is: . S\:=\:\frac{\frac{12}{100}}{1 - \frac{1}{100}}\:=\:\frac{\frac{12}{100}}{\frac{99}  {100}} \:=\:\frac{12}{99} \:=\:\boxed{\frac{4}{33}}



    Method #2: .Let x \:=\:0.12121212\hdots

    Multiply by 100: . 100x \:=\:12.12121212\hdots
    . . . Subtract x:\qquad x \:=\:\;\;0.12121212\hdots

    . .And we have: . . 99x \:=\:12\quad\Rightarrow\quad x \:=\:\frac{12}{99} \:=\:\boxed{\frac{4}{33}}

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  3. #3
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    Talking thank you....

    thanks! it totally makes sense!
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  4. #4
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    Red face can you...

    can you possiby hepl me with my first question? thanks!
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  5. #5
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    Please Help!!!

    could you possibly help me out? I am stuck on two questions
    the general term of an infinite sereis is tn= ( pie square) (3/11)^ n+1. how many terms must be added to reach a sum that is greater than 0.32?
    t1 = (pie square) (3/11)2 = 0.233672181
    r = (3/11)
    Sm = a * (1-rm)/(1-r)
    0.32 * 8/11 = 0.233672181 * (1-rm)
    I know I have to use logarithm but I get confused!
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  6. #6
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    Lexington, MA (USA)
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    Hello, Andrew!

    There must a typo in the first problem . . .


    the general term of an infinite sereis is tn= ( pie square) (3/11)^ n+1.
    how many terms must be added to reach a sum that is greater than 0.32?

    It says: . t_n \:=\:\pi^2\left(\frac{3}{11}\right)^{n+1}

    But the first term: t_1\:=\:\pi^2\left(\frac{3}{11}\right)^2 \:=\:0.734...
    . . is already greater tha 0.32.


    t1 = (pie square) (3/11)2 = 0.233672181

    Here you used just \pi, not \pi^2.
    So I'll assume that the "square" is a typo.
    [By the way, it's not "pie" . . . it's "pi".]


    We have: . t_n \:=\:\pi\left(\frac{3}{11}\right)^{n+1} . . . and: . t_1\:=\:\pi\left(\frac{3}{11}\right)^2\:=\:\frac{9  \pi}{121}

    Sum of a geometric series: . S_m \:=\:a\cdot\frac{1 - r^m}{1 - r}

    So we have: . S_m\:=\:\frac{9\pi}{121}\cdot\frac{1 - \left(\frac{3}{11}\right)^m}{1 - \frac{3}{11}} \:=\:\frac{9\pi}{88}\left[1 - \left(\frac{3}{11}\right)^m\right]


    We want this sum to exceed 0.32, so we have:
    . . \frac{9\pi}{88}\left[1 - \left(\frac{3}{11}\right)^m\right]\;>\;0.32 . . . and we must solve for m.

    Divide by \frac{9\pi}{88}:\;\;1 - \left(\frac{3}{11}\right)^m \;> \:0.995956266

    Subtract 1: . -\left(\frac{3}{11}\right)^m \;> \:-0.004043724

    Multiply by -1:\;\;\left(\frac{3}{11}\right)^m \:<\:0.004043734 . . .
    reverse the inequality!

    Take logs: . \log\left(\frac{3}{11}\right)^m \:< \:\log(0.004043724)

    . . . . . . . m\cdot\log\left(\frac{3}{11}\right) \;< \;\log(0.004043724)

    Hence:. . . . . . . . . . m \;> \;\frac{\log(0.004043734)}{\log(3/11)} \;=\;4.241252175
    . .
    Note: log(3/11) is negative, so we must reverse the inequality again.


    Therefore: . \boxed{m \,=\,5}

    For the sum to exceed 0.32, add the first five terms.

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