log help needed

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  1. dongonaeatu's Avatar
    1 12-04-2012  20:05
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    log help needed
    write 3log5-log2 as a single logarithm

    do i do 5log3-log2
    so 15log-log2

    so log7.5? is this right
  2. TGH1's Avatar
    2 12-04-2012  20:10
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    Re: log help needed
    (Original post by dongonaeatu)
    write 3log5-log2 as a single logarithm

    do i do 5log3-log2
    so 15log-log2

    so log7.5? is this right
    No, as you haven't applied the log rules correctly.

    alnb = ln(b^a)
    lnc - lnd = ln(c/d)
  3. raheem94's Avatar
    3 12-04-2012  20:11
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    Re: log help needed
    (Original post by dongonaeatu)
    write 3log5-log2 as a single logarithm

    do i do 5log3-log2
    so 15log-log2

    so log7.5? is this right
    Remember, \displaystyle alogb=logb^a \ and \ logb-loga=log\left(\frac{b}{a}\right)
  4. dongonaeatu's Avatar
    4 12-04-2012  20:11
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    Re: log help needed
    (Original post by TGH1)
    No, as you haven't applied the log rules correctly.

    alnb = ln(b^a)
    lnc - lnd = ln(c/d)
    is the answer log 62.5
  5. raheem94's Avatar
    5 12-04-2012  20:11
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    Re: log help needed
    (Original post by TGH1)
    No, as you haven't applied the log rules correctly.

    alnb = ln(b^a)
    lnc - lnd = ln(c/d)
    The OP looks to have been stuck on a C2 question so it was better not to use the natural log.
  6. dongonaeatu's Avatar
    6 12-04-2012  20:11
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    Re: log help needed
    (Original post by raheem94)
    Remember, \displaystyle alogb=logb^a \ and \ logb-loga=log\left(\frac{b}{a}\right)
    is the answer log 62.5
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  7. raheem94's Avatar
    7 12-04-2012  20:12
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    Re: log help needed
    (Original post by dongonaeatu)
    is the answer log 62.5
    Yeah
    Post rating:
    2
  8. The Mr Z's Avatar
    8 12-04-2012  20:14
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    Re: log help needed
    (Original post by dongonaeatu)
    is the answer log 62.5
    No, 5log3 = log(3^5) = log(243)

    you have done log(5^3) which is incorrect.
  9. dongonaeatu's Avatar
    9 12-04-2012  20:15
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    Re: log help needed
    (Original post by raheem94)
    Yeah
    sorry to keep bothering you.
    if log10 a=4.1 find the value of a (1mark)

    is it just a=4.1/log10?
  10. dongonaeatu's Avatar
    10 12-04-2012  20:16
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    Re: log help needed
    (Original post by The Mr Z)
    No, 5log3 = log(3^5) = log(243)

    you have done log(5^3) which is incorrect.
    the question was 3log5 not 5log3
  11. The Mr Z's Avatar
    11 12-04-2012  20:19
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    Re: log help needed
    (Original post by dongonaeatu)
    the question was 3log5 not 5log3
    good point, in which case also check the second line of your first post - it's good to be consistent

    then log 62.5 is correct
  12. raheem94's Avatar
    12 12-04-2012  20:19
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    Re: log help needed
    (Original post by dongonaeatu)
    sorry to keep bothering you.
    if log10 a=4.1 find the value of a (1mark)

    is it just a=4.1/log10?
    Remember, \displaystyle log_ab=x \implies a^{x} = b
  13. dongonaeatu's Avatar
    13 12-04-2012  20:22
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    Re: log help needed
    (Original post by raheem94)
    Remember, \displaystyle log_ab=x \implies a^{x} = b
    10^4.1=a?

    or is it log10^4.1=a
  14. raheem94's Avatar
    14 12-04-2012  20:23
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    Re: log help needed
    (Original post by dongonaeatu)
    10^4.1=a?
    Correct.
  15. dongonaeatu's Avatar
    15 12-04-2012  20:25
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    Re: log help needed
    (Original post by raheem94)
    Correct.
    if i do 10^4.1= 12589.25412

    if i do log10^4.1= 0.6127838567 which one is it
  16. raheem94's Avatar
    16 12-04-2012  20:27
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    Re: log help needed
    (Original post by dongonaeatu)
    if i do 10^4.1= 12589.25412

    if i do log10^4.1= 0.6127838567 which one is it
    \displaystyle log_{10}a=4.1 \implies 10^{4.1} = a = 12589.25412
  17. dongonaeatu's Avatar
    17 12-04-2012  20:28
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    Re: log help needed
    (Original post by raheem94)
    \displaystyle log_{10}a=4.1 \implies 10^{4.1} = a = 12589.25412
    thanks
  18. raheem94's Avatar
    18 12-04-2012  20:29
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    Re: log help needed
    (Original post by dongonaeatu)
    thanks
    No problem
  19. dongonaeatu's Avatar
    19 12-04-2012  20:30
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    Re: log help needed
    (Original post by raheem94)
    No problem
    evaluate log3 1/9.... im pretty sure this is -2? as 3^2 is 9 and because its 1/9 its am minus so am i correct in saying the answer is -2
  20. raheem94's Avatar
    20 12-04-2012  20:33
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    Re: log help needed
    (Original post by dongonaeatu)
    evaluate log3 1/9.... im pretty sure this is -2? as 3^2 is 9 and because its 1/9 its am minus so am i correct in saying the answer is -2
    \displaystyle log_3\left(\frac19\right) = log_33^{-2} = -2log_33 = -2 \times 1=-2
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