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Thread: Solving for n

  1. #1
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    Solving for n

    The numbers log (a^3b^7), log(a^5,b^12) and log(a^8,b^15), are the first three terms of an arithmetic sequence, and the the 12th term of the sequence is log(b^n), find the balue of n.
    Possible choices of answers.
    a) 40
    b)56
    c)76
    d)112
    e)143
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  2. #2
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    Hello Tiger
    Quote Originally Posted by Tiger View Post
    The numbers log (a^3b^7), log(a^5,b^12) and log(a^8,b^15), are the first three terms of an arithmetic sequence, and the the 12th term of the sequence is log(b^n), find the balue of n.
    Possible choices of answers.
    a) 40
    b)56
    c)76
    d)112
    e)143
    The first three terms are:
    $\displaystyle \log(a^3b^7),\; \log(a^5b^{12})$ and $\displaystyle \log(a^8b^{15})$
    i.e.
    $\displaystyle 3\log(a) + 7 \log(b),\;5\log(a) + 12\log(b)$ and $\displaystyle 8\log(a)+15\log(b)$
    The difference between consecutive terms is constant. So:
    $\displaystyle 2\log(a)+5\log(b) = 3\log(a)+3\log(b)$

    $\displaystyle \Rightarrow 2\log(b) = \log(a)$
    So the common difference is $\displaystyle 2(2\log(b))+5\log(b)=9\log(b)$.

    The twelfth term is, therefore:
    $\displaystyle 3\log(a)+7\log(b) + 11\times 9 \log(b)$
    $\displaystyle = 6\log(b) + 7\log(b) + 99\log(b) $

    $\displaystyle = 112\log(b)$

    $\displaystyle =\log\big(b^{112}\big)$
    $\displaystyle \Rightarrow n = 112$
    Grandad
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  3. #3
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    Hello, Tiger!

    Great solution, Grandad!
    Here's my approach . . .


    The numbers $\displaystyle \log(a^3b^7),\;\log(a^5b^{12}),\;\log(a^8b^{15})$
    . . are the first three terms of an arithmetic sequence.
    The the 12th term of the sequence is $\displaystyle \log(b^n)$

    Find the value of $\displaystyle n.$

    . . $\displaystyle (a)\;40 \qquad (b)\;56 \qquad (c)\;76 \qquad (d)\;112 \qquad (e)\;143$

    We have: .$\displaystyle \begin{Bmatrix}
    t_1 &=& \log(a^3b^7) \\ \\[-4mm] t_2 &=& \log(a^5b^{12}) \\ \\[-4mm]
    t_3 &=& \log(a^8b^{15}) \end{Bmatrix}\quad [1]$

    $\displaystyle d \:=\:t_2-t_1 \:=\:\log(a^5b^{12}) - \log(a^3b^7) \:=\: \log\left(\frac{a^5b^{12}}{a^3b^7}\right) \:=\:\log(a^2b^5)\;\;[2]$

    $\displaystyle d \:=\:t_3-t_2 \:=\:\log(a^8b^{15}) - \log(a^5b^{12}) \:=\:\log\left(\frac{a^8b^{15}}{a^5b^{12}}\right) \:=\:\log(a^3b^3)\;\;[3] $


    Equate [3] and [2]: .$\displaystyle \log(a^3b^3) \:=\:\log(a^2b^5) \quad\Rightarrow\quad a^3b^3 \:=\:a^2b^5 \quad\Rightarrow\quad a \:=\:b^2$


    Substitute into [1]: .$\displaystyle \begin{Bmatrix}t_1 &=& \log(b^{13}) \\ t_2 &=& \log(b^{22}) \\ t_3 &=& \log(b^{31}) \end{Bmatrix}$

    Substitute into [2]: .$\displaystyle d \:=\:\log(b^9)$


    The 12th term is: .$\displaystyle t_{12} \:=\:t_1 + 11d$

    We have: .$\displaystyle t_{12} \;\;=\;\;\log(b^{13}) + 11\!\cdot\!\log(b^9) \;\;=\;\;\log(b^{13}) + \log(b^{99}) \;\;=\;\;\log(b^{112})$


    Therefore: .$\displaystyle \log(b^n) \:=\:\log(b^{112}) \quad\Rightarrow\quad n \:=\:112$

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