geometric progession help

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  1. TenOfThem's Avatar
    21 12-04-2012  21:53
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    Re: geometric progession help
    no

    3 * 5^x = 150

    5^x = 50

    log5^x = log50

    xlog5 = log50

    x = log50/log5


    Can I ask if you are self teaching this material ... if not, there seems to be a lot that you have not understood from lessons
  2. TenOfThem's Avatar
    22 12-04-2012  21:53
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    Re: geometric progession help
    (Original post by dongonaeatu)
    is the sum to infinity 16.6
    as my post said ... NO
  3. raheem94's Avatar
    23 12-04-2012  21:54
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    Re: geometric progession help
    (Original post by dongonaeatu)
    hey man, solve 3*5^x=150 giving it to 4dp

    is it; log3*5^x=log150

    5^x=log150/log3

    5^x=4.560876795 then i dont know how to get the x on its own
    \displaystyle 3 \times 5^{x} = 150 \implies 5^{x} = 50

    Now take log of both sides and remember, \displaystyle logb^a = alogb
  4. dongonaeatu's Avatar
    24 12-04-2012  21:57
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    Re: geometric progession help
    (Original post by raheem94)
    \displaystyle 3 \times 5^{x} = 150 \implies 5^{x} = 50

    Now take log of both sides and remember, \displaystyle logb^a = alogb
    oh i see so you get rid of the 3* so its 5^x=50 then do i do

    log5^x=log50
  5. raheem94's Avatar
    25 12-04-2012  21:58
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    Re: geometric progession help
    (Original post by dongonaeatu)
    oh i see so you get rid of the 3* so its 5^x=50 then do i do

    log5^x=log50
    Yes
  6. dongonaeatu's Avatar
    26 12-04-2012  22:02
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    Re: geometric progession help
    (Original post by TenOfThem)
    no

    3 * 5^x = 150

    5^x = 50

    log5^x = log50

    xlog5 = log50

    x = log50/log5


    Can I ask if you are self teaching this material ... if not, there seems to be a lot that you have not understood from lessons
    thanks that was helpful, is the answer 2.4307? and no i do AS maths at college
  7. raheem94's Avatar
    27 12-04-2012  22:02
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    Re: geometric progession help
    (Original post by dongonaeatu)
    oh i see so you get rid of the 3* so its 5^x=50 then do i do

    log5^x=log50
    By the way, you need to learn the log chapter again. Going through your book will be more helpful than asking questions here.

    I am quite sure if you learn the chapter well, then you should be able to answer most of these questions.
  8. raheem94's Avatar
    28 12-04-2012  22:03
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    Re: geometric progession help
    (Original post by dongonaeatu)
    thanks that was helpful, is the answer 2.4307? and no i do AS maths at college
    Yes
  9. dongonaeatu's Avatar
    29 12-04-2012  22:03
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    Re: geometric progession help
    (Original post by raheem94)
    By the way, you need to learn the log chapter again. Going through your book will be more helpful than asking questions here.

    I am quite sure if you learn the chapter well, then you should be able to answer most of these questions.
    thanks man i will do that
  10. dongonaeatu's Avatar
    30 12-04-2012  22:11
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    Re: geometric progession help
    (Original post by raheem94)
    Yes
    The first term of a geometric progression is 5 and the third term is 10
    i. determine the two possible values for the common ratio [2marks]

    so a=5 and ar is the second term so ar^2=10

    so 5(r)^2=10

    r^2=5 so r=square root of 5 so thats 2.236067977 or -2.236067977

    is this right
    Last edited by dongonaeatu; 12-04-2012 at 22:15.
  11. dongonaeatu's Avatar
    31 12-04-2012  22:16
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    Re: geometric progession help
    (Original post by raheem94)
    Yes
    sorry that was wrong

    i think its: ar^2=10
    5(r)^2=10
    so then i divide by 5 NOT minus 5 like i did last time becauses it's 5 times

    so r^2=2
    so r= square root of 2 which equals 1.414213562 or -1.414213562 please tell me this is correct
  12. raheem94's Avatar
    32 12-04-2012  22:19
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    Re: geometric progession help
    (Original post by dongonaeatu)
    sorry that was wrong

    i think its: ar^2=10
    5(r)^2=10
    so then i divide by 5 NOT minus 5 like i did last time becauses it's 5 times

    so r^2=2
    so r= square root of 2 which equals 1.414213562 or -1.414213562 please tell me this is correct
    It is correct
  13. dongonaeatu's Avatar
    33 12-04-2012  22:29
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    Re: geometric progession help
    (Original post by raheem94)
    It is correct
    yay!!

    ii. Using the largest of these two values for the common ratio find the first term to exceed 5000. [3 marks]

    so i am using the positive 1.414213562 but i really dont know how to tackle this type of question
  14. raheem94's Avatar
    34 12-04-2012  22:34
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    Re: geometric progession help
    (Original post by dongonaeatu)
    yay!!

    ii. Using the largest of these two values for the common ratio find the first term to exceed 5000. [3 marks]

    so i am using the positive 1.414213562 but i really dont know how to tackle this type of question
    You know that nth term = ar^{n-1}

    So here, 5000=5r^{n-1}

    Find the value of 'n'.
  15. dongonaeatu's Avatar
    35 12-04-2012  22:38
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    Re: geometric progession help
    (Original post by raheem94)
    You know that nth term = ar^{n-1}

    So here, 5000=5r^{n-1}

    Find the value of 'n'.
    so 5000= 5(1.414213562)^n-1

    5000=7.07106701^n-1

    how do i get n
  16. Joshmeid's Avatar
    36 12-04-2012  22:41
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    Re: geometric progession help
    (Original post by dongonaeatu)
    so 5000= 5(1.414213562)^n-1

    5000=7.07106701^n-1

    how do i get n
    Take the log of both sides

    Also to make it easier, keep it in root form

    As in, instead of 1.414213562, use \sqrt 2
    Last edited by Joshmeid; 12-04-2012 at 22:44.
  17. dongonaeatu's Avatar
    37 12-04-2012  22:42
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    Re: geometric progession help
    (Original post by Joshmeid)
    Take the log of both sides
    this isnt a log question its geometric progression
  18. raheem94's Avatar
    38 12-04-2012  22:43
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    Re: geometric progession help
    (Original post by dongonaeatu)
    so 5000= 5(1.414213562)^n-1

    5000=7.07106701^n-1

    how do i get n
    You really need to go through the chapter again.

    \displaystyle 5000=5(\sqrt2)^{n-1} \implies 1000=(\sqrt2)^{n-1} \implies log1000=log(\sqrt2)^{n-1} \implies log1000=(n-1)log\sqrt2 \implies log1000=nlog\sqrt2 - log\sqrt2 \implies \frac{log1000+log\sqrt2}{log \sqrt2}=n \implies n=20.9315

    The answer is \boxed{n=21}
  19. dongonaeatu's Avatar
    39 12-04-2012  22:45
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    Re: geometric progession help
    (Original post by raheem94)
    You really need to go through the chapter again.

    \displaystyle 5000=5(\sqrt2)^{n-1} \implies 1000=(\sqrt2)^{n-1} \implies log1000=log(\sqrt2)^{n-1} \implies log1000=(n-1)log\sqrt2 \implies log1000=nlog\sqrt2 - log\sqrt2 \implies \frac{log1000+log\sqrt2}{log \sqrt2}=n \implies n=20.9315

    The answer is \boxed{n=21}
    thats insane, and i never knew logs were used in geometric progressions; how did you know to use logs
  20. raheem94's Avatar
    40 12-04-2012  22:48
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    Re: geometric progession help
    (Original post by dongonaeatu)
    thats insane, and i never knew logs were used in geometric progressions; how did you know to use logs
    The only way to solve this was to use logs, as said before, you need to go through your book again.
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