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Math Help - Finding nth term

  1. #1
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    Finding nth term

    Sequence: log(subscript)2) 8, log(subscript)4) 8, log(subscript)8) 8, log(subscript)16) 8, log(subscript)32) 8, log(subscript)64) 8…..
    How do I find an expression for the nth term of the sequence in the form p/q, where p and q are integers.
    I would appreciate your help...
    Have an awesome day!!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by vale01 View Post
    Sequence: log(subscript)2) 8, log(subscript)4) 8, log(subscript)8) 8, log(subscript)16) 8, log(subscript)32) 8, log(subscript)64) 8…..
    How do I find an expression for the nth term of the sequence in the form p/q, where p and q are integers.
    Two things that could be helpful :
    • \log_b x = \frac{\ln x}{\ln b}
    • \ln x^y = y\ln x
    Using this you can write each term of the sequence as a fraction. For example, for the fourth term :

    <br />
\log_{16} 8=\frac{\ln 8}{\ln 16}=\frac{\ln 2^3}{\ln 2^4}=\frac{3\ln 2}{4\ln 2}=\frac34
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  3. #3
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    thanks

    thank u so much!
    that helps alot
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  4. #4
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    Hello, vale01!

    Sequence: . \log_2(8),\;\log_4(8),\; \log_8(8),\;\log_{16}(8),\;\log_{32}(8),\;\log_{64  }(8),\;\hdots

    Find an expression for the n^{th} term in the form \frac{p}{q}, where p and q are integers.
    Evaluate the logs ... Look for a pattern.

    \log_2(8) = x\quad\Rightarrow\quad 2^x = 8 \quad\Rightarrow\quad 2^x = 2^3 \quad\Rightarrow\quad x = 3

    \log_4(8) = x \quad\Rightarrow\quad 4^x = 8\quad\Rightarrow\quad (2^2)^x = 2^3\quad\Rightarrow\quad 2x = 3 \quad\Rightarrow\quad x = \frac{3}{2}

    \log_8(8) = x \quad\Rightarrow\quad 8^x = 8 \quad\Rightarrow\quad x = 1

    \log_{16}(8) = x \quad\Rightarrow\quad 16^x = 8 \quad\Rightarrow\quad (2^4)^x = 2^3 \quad\Rightarrow\quad 4x = 3 \quad\Rightarrow\quad x = \frac{3}{4}

    \log_{32}(8) = x \quad\Rightarrow\quad 32^x = 8 \quad\Rightarrow\quad (2^5)^x = 2^3 \quad\Rightarrow\quad 5x = 3 \quad\Rightarrow\quad x = \frac{3}{5}

    \log_{64}(8) = x \quad\Rightarrow\quad 64^x = 8 \quad\Rightarrow\quad (2^6)^x = 2^3 \quad\Rightarrow\quad 6x = 3 \quad\Rightarrow\quad x = \frac{1}{2}


    \begin{array}{c|ccccccc}<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & \hdots \\ \hline \hline \\[-4mm]<br />
\text{We have:}& 3 & \frac{3}{2} & 1 & \frac{3}{4} & \frac{3}{5} & \frac{1}{2} & \hdots \\ \\[-4mm] \hline \\[-4mm]<br />
\text{That is:} & 3 & \frac{3}{2} & \frac{3}{3} & \frac{3}{4} & \frac{3}{5} & \frac{3}{6} & \hdots \\ \\[-4mm] \hline<br />
\end{array}

    And we see that the n^{th} term is: . a_n \:=\:\frac{3}{n}

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  5. #5
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    thanks

    thank u!
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