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Math Help - logarithms and geometric sequences

  1. #1
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    logarithms and geometric sequences

    i was going through my hw and got stuck on this question- and i haven't been able to figure how on earth to do it! does anyone have any ideas? this is really bugging me now.

    if log2 x, (1 + log4 x), log8 4x are consecutive terms of a geometric sequence, determine the possible values of x.

    all i know is the properties of a logarithm and the formula for a geometric sequence is tn= ar^(n-1). is it possible for me to use the concept of t2/t1= t3/t2, etc with logs? and how can i find x? i really dont have a clue and id be happy for any hints, advice or an answer.
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    use

    log_b(x) = \frac{log_a(x)}{log_a(b)}

    to make log_2(x) and log_8(4x)

    into base 4.
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  3. #3
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    Quote Originally Posted by dark flame View Post
    i was going through my hw and got stuck on this question- and i haven't been able to figure how on earth to do it! does anyone have any ideas? this is really bugging me now.

    if log2 x, (1 + log4 x), log8 4x are consecutive terms of a geometric sequence, determine the possible values of x.

    all i know is the properties of a logarithm and the formula for a geometric sequence is tn= ar^(n-1). is it possible for me to use the concept of t2/t1= t3/t2, etc with logs? and how can i find x? i really dont have a clue and id be happy for any hints, advice or an answer.
    If t_1 = \log_2{x}, t_2 = 1 + \log_4{x}, t_3 = \log_8{x} are terms of a geometric sequence, then

    \frac{t_3}{t_2} = \frac{t_2}{t_1}

    \frac{\log_8{(4x)}}{1 + \log_4{x}} = \frac{1 + \log_4{x}}{\log_2{x}}

    \log_8{(4x)}\log_2{x} = (1 + \log_4{x})^2.


    It helps to use the Change of base formula here

    \log_b{x} = \frac{\log_k{x}}{\log_k{b}}


    So \frac{\ln{(4x)}}{\ln{8}}\,\frac{\ln{x}}{\ln{2}} = \left(1 + \frac{\ln{x}}{\ln{4}}\right)^2

    I trust you'll be able to go from here...
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    logarithms with e

    Quote Originally Posted by Prove It View Post
    If t_1 = \log_2{x}, t_2 = 1 + \log_4{x}, t_3 = \log_8{x} are terms of a geometric sequence, then

    \frac{t_3}{t_2} = \frac{t_2}{t_1}

    \frac{\log_8{(4x)}}{1 + \log_4{x}} = \frac{1 + \log_4{x}}{\log_2{x}}

    \log_8{(4x)}\log_2{x} = (1 + \log_4{x})^2.


    It helps to use the Change of base formula here

    \log_b{x} = \frac{\log_k{x}}{\log_k{b}}


    So \frac{\ln{(4x)}}{\ln{8}}\,\frac{\ln{x}}{\ln{2}} = \left(1 + \frac{\ln{x}}{\ln{4}}\right)^2

    I trust you'll be able to go from here...
    the thing is that we've only briefly went other the "ln" and "e" thing for maybe not even 7 minutes- i dont quite know or understand when or how to use it unfortunetly... ive only ever used base 10.
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    how exactly do i use the "ln" and when do i know how to use it?
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    Quote Originally Posted by dark flame View Post
    the thing is that we've only briefly went other the "ln" and "e" thing for maybe not even 7 minutes- i dont quite know or understand when or how to use it unfortunetly... ive only ever used base 10.
    When you use a Change of Base rule, you can change the base to whatever you like.

    \log_{10} is fine, as is any other base.

    I just naturally choose \ln because it is the most commonly used.

    Once you've converted everything to the same base, you use your standard logarithm and algebra rules to simplify and solve for x.
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  7. #7
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    Quote Originally Posted by dark flame View Post
    i was going through my hw and got stuck on this question- and i haven't been able to figure how on earth to do it! does anyone have any ideas? this is really bugging me now.

    if log2 x, (1 + log4 x), log8 4x are consecutive terms of a geometric sequence, determine the possible values of x.

    all i know is the properties of a logarithm and the formula for a geometric sequence is tn= ar^(n-1). is it possible for me to use the concept of t2/t1= t3/t2, etc with logs? and how can i find x? i really dont have a clue and id be happy for any hints, advice or an answer.
    {\text{Since they form a geometric sequence,}} \hfill \\

     \frac{{t_2 }}<br />
{{t_1 }} = \frac{{t_3 }}<br />
{{t_2 }} \hfill \\

     \Rightarrow \frac{{1 + \log _4 x}}<br />
{{\log _2 x}} = \frac{{\log _8 4x}}<br />
{{1 + \log _4 x}} \hfill \\

     \Rightarrow \left( {1 + \log _4 x} \right)^2  = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\

     \Rightarrow \left( {\log _4 4 + \log _4 x} \right)^2  = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\

    \Rightarrow \left( {\log _4 4x} \right)^2  = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\

       \Rightarrow \left( {\frac{{\ln 4x}}<br />
{{\ln 4}}} \right)^2  = \left( {\frac{{\ln x}}<br />
{{\ln 2}}} \right)\left( {\frac{{\ln 4x}}<br />
{{\ln 8}}} \right) \hfill \\

      \Rightarrow \left( {\frac{{\ln 4x}}<br />
{{2\ln 2}}} \right)^2  = \left( {\frac{{\ln x}}<br />
{{\ln 2}}} \right)\left( {\frac{{\ln 4x}}<br />
{{3\ln 2}}} \right) \hfill \\

    \Rightarrow \frac{1}<br />
{4}\left( {\ln 4x} \right)^2  = \frac{1}<br />
{3}\left( {\ln x} \right)\left( {\ln 4x} \right) \hfill \\

     \Rightarrow 3\left( {\ln 4x} \right)^2  = 4\left( {\ln x} \right)\left( {\ln 4x} \right) \hfill \\

     \Rightarrow 3\left( {\ln 4x} \right)^2  - 4\left( {\ln x} \right)\left( {\ln 4x} \right) = 0 \hfill \\

     \Rightarrow \ln 4x\left( {3\ln 4x - 4\ln x} \right) = 0 \hfill \\

     \Rightarrow \ln 4x = 0 \Rightarrow 4x = 1 \Rightarrow x = \frac{1}<br />
{4} \hfill \\

     {\text{OR}} \hfill \\

     3\ln 4x - 4\ln x = 0 \hfill \\

    3\left( {\ln 4 + \ln x} \right) - 4\ln x = 0 \hfill \\

     3\ln 4 - \ln x = 0 \hfill \\

    \ln 4^3  = \ln x \hfill \\

    x = 64 \hfill \\

    Check the answers by putting these values of x, do they form a geometric sequence?
    Last edited by Shyam; May 5th 2009 at 05:50 PM.
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  8. #8
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    Quote Originally Posted by dark flame View Post
    how exactly do i use the "ln" and when do i know how to use it?
    ln(x) = log_e(x)
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  9. #9
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    alittle confused

    Quote Originally Posted by Shyam View Post
    {\text{Since they form a geometric sequence,}} \hfill \\

     \frac{{t_2 }}<br />
{{t_1 }} = \frac{{t_3 }}<br />
{{t_2 }} \hfill \\

     \Rightarrow \frac{{1 + \log _4 x}}<br />
{{\log _2 x}} = \frac{{\log _8 4x}}<br />
{{1 + \log _4 x}} \hfill \\

     \Rightarrow \left( {1 + \log _4 x} \right)^2 = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\

     \Rightarrow \left( {\log _4 4 + \log _4 x} \right)^2 = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\

    \Rightarrow \left( {\log _4 4x} \right)^2 = \left( {\log _2 x} \right)\left( {\log _8 4x} \right) \hfill \\

     \Rightarrow \left( {\frac{{\ln 4x}}<br />
{{\ln 4}}} \right)^2 = \left( {\frac{{\ln x}}<br />
{{\ln 2}}} \right)\left( {\frac{{\ln 4x}}<br />
{{\ln 8}}} \right) \hfill \\

     \Rightarrow \left( {\frac{{\ln 4x}}<br />
{{2\ln 2}}} \right)^2 = \left( {\frac{{\ln x}}<br />
{{\ln 2}}} \right)\left( {\frac{{\ln 4x}}<br />
{{3\ln 2}}} \right) \hfill \\

    \Rightarrow \frac{1}<br />
{4}\left( {\ln 4x} \right)^2 = \frac{1}<br />
{3}\left( {\ln x} \right)\left( {\ln 4x} \right) \hfill \\

     \Rightarrow 3\left( {\ln 4x} \right)^2 = 4\left( {\ln x} \right)\left( {\ln 4x} \right) \hfill \\

     \Rightarrow 3\left( {\ln 4x} \right)^2 - 4\left( {\ln x} \right)\left( {\ln 4x} \right) = 0 \hfill \\

     \Rightarrow \ln 4x\left( {3\ln 4x - 4\ln x} \right) = 0 \hfill \\

     \Rightarrow \ln 4x = 0 \Rightarrow 4x = 1 \Rightarrow x = \frac{1}<br />
{4} \hfill \\

     {\text{OR}} \hfill \\

     3\ln 4x - 4\ln x = 0 \hfill \\

    3\left( {\ln 4 + \ln x} \right) - 4\ln x = 0 \hfill \\

     3\ln 4 - \ln x = 0 \hfill \\

    \ln 4^3 = \ln x \hfill \\

    x = 64 \hfill \\

    Check the answers by putting these values of x, do they form a geometric sequence?
    i was going through some of what you did originally before using natural logs and i got stuck at the point whree you got:

    1/4 (log 4 x)^2 = 1/3 (log x) (log 4x) i just dont quite understand your reasoning behind this- everything else before this makes perfect sense to me.
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