Results are out! Find what you need...fast. Get quick advice or join the chat
Hey there Sign in to join this conversationNew here? Join for free

geometric progession help

Announcements Posted on
    • 58 followers
    Online

    ReputationRep:
    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
    • 58 followers
    Online

    ReputationRep:
    (Original post by dongonaeatu)
    is the sum to infinity 16.6
    as my post said ... NO
    • 46 followers
    Offline

    ReputationRep:
    (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
    • Thread Starter
    • 5 followers
    Offline

    ReputationRep:
    (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
    • 46 followers
    Offline

    ReputationRep:
    (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
    • Thread Starter
    • 5 followers
    Offline

    ReputationRep:
    (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
    • 46 followers
    Offline

    ReputationRep:
    (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.
    • 46 followers
    Offline

    ReputationRep:
    (Original post by dongonaeatu)
    thanks that was helpful, is the answer 2.4307? and no i do AS maths at college
    Yes
    • Thread Starter
    • 5 followers
    Offline

    ReputationRep:
    (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
    • Thread Starter
    • 5 followers
    Offline

    ReputationRep:
    (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
    • Thread Starter
    • 5 followers
    Offline

    ReputationRep:
    (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
    • 46 followers
    Offline

    ReputationRep:
    (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
    • Thread Starter
    • 5 followers
    Offline

    ReputationRep:
    (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
    • 46 followers
    Offline

    ReputationRep:
    (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'.
    • Thread Starter
    • 5 followers
    Offline

    ReputationRep:
    (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
    • 0 followers
    Offline

    ReputationRep:
    (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
    • Thread Starter
    • 5 followers
    Offline

    ReputationRep:
    (Original post by Joshmeid)
    Take the log of both sides
    this isnt a log question its geometric progression
    • 46 followers
    Offline

    ReputationRep:
    (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}
    • Thread Starter
    • 5 followers
    Offline

    ReputationRep:
    (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
    • 46 followers
    Offline

    ReputationRep:
    (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.

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: April 12, 2012
New on TSR

Student in a Million awards

All the results from our prize-giving night

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.