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Math Help - Logarithm

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    10

    Logarithm

    hey, i have a question and also the answers,
    ive worked it out and have the first answer but i just get my head round to how the 2nd answer is....

    this is the q..
    given that y=ax^n + 3, that y=4.4 when x=10 and y=12.8 when x=100, find the value of 'n' and 'a'

    the 2 answers are 0.845 & 0.2

    this is my working out for the first answer....

    y = ax^n +3 => y=4.4, x=10

    4.4 = a10^n + 3

    a = (4.4 - 3) / 10^n

    = 1.4/10^n

    y = 12.8, x = 100

    12.8 = a100^n + 3

    12.8 = (1.4/10^n) x 100^n + 3

    so 100^n = 10^2n

    12.8 = (1.4/10^n) x 10^2n + 3

    9.8 = 1.4 x 10^n

    10^n = 7

    log7 = log10^n

    n = (log7)/(log10)

    = 0.845




    i am so sorry if this is long, im currently in 6th form, studying IB
    & in order for me to learn, i like having the steps in my working out rite...



    so how would i get to the answer 0.2?
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  2. #2
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by exclusiive View Post
    hey, i have a question and also the answers,
    ive worked it out and have the first answer but i just get my head round to how the 2nd answer is....

    this is the q..
    given that y=ax^n + 3, that y=4.4 when x=10 and y=12.8 when x=100, find the value of 'n' and 'a'

    the 2 answers are 0.845 & 0.2

    this is my working out for the first answer....

    y = ax^n +3 => y=4.4, x=10

    4.4 = a10^n + 3

    a = (4.4 - 3) / 10^n

    = 1.4/10^n

    y = 12.8, x = 100

    12.8 = a100^n + 3

    12.8 = (1.4/10^n) x 100^n + 3

    so 100^n = 10^2n

    12.8 = (1.4/10^n) x 10^2n + 3

    9.8 = 1.4 x 10^n

    10^n = 7

    log7 = log10^n

    n = (log7)/(log10)

    = 0.845




    i am so sorry if this is long, im currently in 6th form, studying IB
    & in order for me to learn, i like having the steps in my working out rite...



    so how would i get to the answer 0.2?
    You have the value of n, so just substitute it in any equation which contains n and a
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