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Math Help - Logarthim - Finding expression - URGENT PLZZ

  1. #1
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    Logarthim - Finding expression - URGENT PLZZ

    log[base2]8, log[base4]8, log[base8]8, log[base18, log[base32]8, log[base64]8,

    I am given this sequence and i need to find an expression for the nth term in form of p/q, where p and q belongs to z(integer).

    i guess it has to do with the bases
    like log[base2]8
    = log8/log2
    = log8 - log2
    = ??????
    then how to find an expression

    I tried the following:

    1) since the bases form a geometric progression
    i found the common ratio
    which is r =Un/Un-1
    ex: 8/4 = 2

    2) The nth term formula for a geometric progression
    is Un= ar^(n-1)
    where a = 1st term, which is Log(base2)8
    r=common ratio, which is 2
    n=the rank

    so the expression is
    Un= Log(base2)8(2)^n-1
    However its not quite correct
    beacuse if we tested it, when n=7
    to complete the sequence
    the value should be log(base128)8
    NOT Log(base2)8 x 128
    I hope you get where i am going

    I really appreciate ypu help
    THANK YOU!!!!!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GoneCrazy View Post
    log[base2]8, log[base4]8, log[base8]8, log[base18, log[base32]8, log[base64]8,

    I am given this sequence and i need to find an expression for the nth term in form of p/q, where p and q belongs to z(integer).

    i guess it has to do with the bases
    like log[base2]8
    = log8/log2
    = log8 - log2
    = ??????
    then how to find an expression

    I tried the following:

    1) since the bases form a geometric progression
    i found the common ratio
    which is r =Un/Un-1
    ex: 8/4 = 2

    2) The nth term formula for a geometric progression
    is Un= ar^(n-1)
    where a = 1st term, which is Log(base2)8
    r=common ratio, which is 2
    n=the rank

    so the expression is
    Un= Log(base2)8(2)^n-1
    However its not quite correct
    beacuse if we tested it, when n=7
    to complete the sequence
    the value should be log(base128)8
    NOT Log(base2)8 x 128
    I hope you get where i am going

    I really appreciate ypu help
    THANK YOU!!!!!!
    you said the base was a geometric progression. why did you not then make the base the geometric progression? i have trouble seeing your logic. would it not rather be a_n = \log_{2^n}8 for n = 1,~2,~3, \cdots ?
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Jhevon View Post
    you said the base was a geometric progression. why did you not then make the base the geometric progression? i have trouble seeing your logic. would it not rather be a_n = \log_{2^n}8 for n = 1,~2,~3, \cdots ?
    a_{n}=\log_{2^{n}}(8) \Rightarrow 2^{na_{n}}=2^{3}
    .................... \Rightarrow 2^{na_{n}-3}=1
    .................... \Rightarrow na_{n}-3=0
    .................... \Rightarrow a_{n}={\color{red}{\frac{3}{n}}}.

    a_{n}={\color{red}{\frac{n-1}{n}}}{\color{red}{\frac{3}{n-1}}}
    ..._ ={\color{red}{\frac{n-1}{n}a_{n}}}

    I dont think that this is a geometric sequence.
    Or if it is so, I have a big mistake.
    Last edited by bkarpuz; September 29th 2008 at 03:53 AM. Reason: mr fantastic's corrections are prompted.
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  4. #4
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    Quote Originally Posted by bkarpuz View Post
    a_{n}=\log_{2^{n}}(8) \Rightarrow 2^{na_{n}}=2^{3}
    .................... \Rightarrow 2^{na_{n}-3}=1
    .................... \Rightarrow na_{n}-3=0
    .................... \Rightarrow a_{n}=-\frac{3}{n}. *

    [snip]
    Correction to line *: \Rightarrow a_{n}=\frac{3}{n}.
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  5. #5
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    Quote Originally Posted by GoneCrazy View Post
    log[base2]8, log[base4]8, log[base8]8, log[base18, log[base32]8, log[base64]8,

    I am given this sequence and i need to find an expression for the nth term in form of p/q, where p and q belongs to z(integer).
    [snip]
    Regarding the red: You mean log[base16]8, right ....?
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  6. #6
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    Yeh it has nothing to do with geometric progression -my mistake
    i didn't take enough time to think it over before asking

    oooh yeh...i meant 16 not 18 - sorry

    I really didn't get your way - Both bkarpuz and Mr.Fantastic,
    if you can explain it more plzzzz

    ---------------------------------------------------

    Here is my way, i figured out 2 ways:

    2,4,8,16,32,64 can be expressed as 2 raised to a certain power
    Then using these two basic rules, each value in the squence can be simplified:
    1) Log (baseB)A = Log A /Log B

    2) Log a^2 = 2Loga

    example: Log (base4 )8

    = Log 8

    Log 4

    = Log 2^3
    Log 2^2

    3 Log 2
    2 Log 2

    = 3
    2

    The same applies for the rest
    making the following pattern:
    3/1 , 3,2 , 3/3, 3/4, 3/5, 3/6

    Therefore, the nth-term is going to be 3/n

    --------------------------------------------

    2) The sequence is in the form of log[base 2^n] 8

    By using the base rule, we can further simplify it beacome in the form of p/q


    log[base 2^n] 8
    = log[base 2] 8 / log [base 2]2^n
    = log[base 2] 2^3 / log [base 2]2^n

    After crossing out the similar logs you are left with 3/n
    ----------------------------------
    I have anthor part of the question if anyone is intrested
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  7. #7
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    Quote Originally Posted by GoneCrazy View Post
    Yeh it has nothing to do with geometric progression -my mistake
    i didn't take enough time to think it over before asking

    oooh yeh...i meant 16 not 18 - sorry

    I really didn't get your way - Both bkarpuz and Mr.Fantastic,
    if you can explain it more plzzzz

    ---------------------------------------------------

    Here is my way, i figured out 2 ways:

    2,4,8,16,32,64 can be expressed as 2 raised to a certain power
    Then using these two basic rules, each value in the squence can be simplified:
    1) Log (baseB)A = Log A /Log B
    2) Log a^2 = 2Loga

    example: Log (base4 )8

    = Log 8
    Log 4

    = Log 2^3
    Log 2^2

    3 Log 2
    2 Log 2

    = 3
    2

    The same applies for the rest
    making the following pattern:
    3/1 , 3,2 , 3/3, 3/4, 3/5, 3/6

    Therefore, the nth-term is going to be 3/n

    --------------------------------------------

    2) The sequence is in the form of log[base 2^n] 8

    By using the base rule, we can further simplify it beacome in the form of p/q


    log[base 2^n] 8
    = log[base 2] 8 / log [base 2]2^n
    = log[base 2] 2^3 / log [base 2]2^n

    After crossing out the similar logs you are left with 3/n
    ----------------------------------
    I have anthor part of the question if anyone is intrested
    If you have more questions related to this one then you should certainly feel free to ask them.
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