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Math Help - summation-related(?) problem

  1. #1
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    summation-related(?) problem

    Hello,

    I'm not sure if this is the right forum -- I'm not a student so I've not arrived at this problem in a textbook context.

    I want to solve an equation of the form:

    (a^n)b-(a^n-1)c-(a^n-2)c-...-ac-c=0

    where a, b, and c are constants and n is the unknown to be found.

    Any pointers -- whether on the problem or on where to go to pose it again -- would be appreciated.

    Regards,
    Isolde.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by isolde View Post
    Hello,

    I'm not sure if this is the right forum -- I'm not a student so I've not arrived at this problem in a textbook context.

    I want to solve an equation of the form:

    (a^n)b-(a^n-1)c-(a^n-2)c-...-ac-c=0

    where a, b, and c are constants and n is the unknown to be found.




    Any pointers -- whether on the problem or on where to go to post it again -- would be appreciated.


    ....Double Posting is against rules if you want to move your thread report it to mods, anyway you have posted it correctly

    Regards,
    Isolde.
    (a^n)b-(a^{n-1})c-(a^{n-2})c-...-ac-c=0

    a^{n}b = (a^{n-1})c+(a^{n-2})c...+ac+c

    a^{n}b = c[(a^{n-1})+(a^{n-2})...+a+1]


    **Now  [(a^{n-1})+(a^{n-2})...+a+1] is a Geometric Progression

    **It is = 1 \frac{a^{n} - 1}{a-1}


    a^n b = c\times \frac{a^{n} - 1}{a-1}

    a^{n+1}b  - a^n b = c(a^n - 1)

    Put a^n =t

    abt +c = t(b+c)

    t(b+c-ab) = c

    Hence
    <br />
t= \frac{c}{b+c-ab}
    <br />
a^n =  \frac{c}{b+c-ab}

    Hence

    n \ln a = \ln \frac{c}{b+c-ab}

    \implies n = \ln (\frac{c}{ab+ac-a^2b})
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  3. #3
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    Thanks for the solution ADARSH.
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  4. #4
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    I`m new to logarithms, so it`s taken me a while to catch on to what you`re doing here. I still don`t understand the last step.

    From

    <br />
n \ln a = \ln \frac{c}{b+c-ab}<br />

    I assume you go by the intermediate step

    <br />
n = \frac{\ln \frac{c}{b+c-ab}}{\ln a}<br />

    in order to derive

    <br />
n = \ln (\frac{c}{ab+ac-a^2b})<br />

    but as I say, not knowing logs very well, I can`t see how this last step to the solution is made. Presumably by some such identity as

    <br />
\frac{\log_a{b}}{\log_a{c}} = \log_a{\frac{b}{c}}<br />

    but I`m not familiar with this one, and haven`t been able to find it at Wikipedia`s list.

    I have another thought. Would it be possible from

    <br />
n = \frac{\ln \frac{c}{b+c-ab}}{\ln a}<br />

    to derive

    <br />
n = \log_a{\frac{c}{b+c-ab}}<br />

    by the identity

    <br />
\log_a{b} = \frac{\log_c{b}}{\log_b{c}}<br />

    ?

    Thanks again,
    Marc.
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  5. #5
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by isolde View Post
    I`m new to logarithms, so it`s taken me a while to catch on to what you`re doing here. I still don`t understand the last step.

    From

    <br />
n \ln a = \ln \frac{c}{b+c-ab}<br />

    I assume you go by the intermediate step

    <br />
n = \frac{\ln \frac{c}{b+c-ab}}{\ln a}<br />

    in order to derive

    <br />
n = \ln_a {(\frac{c}{b+c-ab})}<br />
    --------------------------------------------------------------
    it was wrong I have corrected it
    You are correct
    You won't find it because it's wrong, Sorry My mistake
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  6. #6
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    Thank-you for your gracious response.

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