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  1. avtar95's Avatar
    1 15-04-2012  01:47
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    Logs
    Hi I don't get this question


    Solve, giving your answer to 3 significant figures:


    log2x+log4x=2

    :d
  2. dslc's Avatar
    2 15-04-2012  01:51
    • Adored and Respected Member
    Re: Logs
    (Original post by avtar95)
    Hi I don't get this question


    Solve, giving your answer to 3 significant figures:


    log2x+log4x=2

    :d
    First of all you need to have the same bases, presuming the 2 and 4 are representing base numbers?

    Are you familiar with the change of base law?
  3. avtar95's Avatar
    3 15-04-2012  01:51
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    Re: Logs
    (Original post by dslc)
    First of all you need to have the same bases.

    Are you familiar with the change of base law?
    could you help
    Post rating:
    1
  4. dslc's Avatar
    4 15-04-2012  01:53
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    Re: Logs
    (Original post by avtar95)
    could you help
    Well is the question log2x+log4x=2?
  5. avtar95's Avatar
    5 15-04-2012  01:54
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    Re: Logs
    (Original post by dslc)
    Well is the question log2x+log4x=2?
    yes
  6. dslc's Avatar
    6 15-04-2012  01:58
    • Adored and Respected Member
    Re: Logs
    (Original post by avtar95)
    yes
    Okay so you can use the change of base law which is: logax = logbx/logba

    So log4x becomes log2x/log24

    Does this help?
  7. dongonaeatu's Avatar
    7 15-04-2012  01:59
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    Re: Logs
    (Original post by avtar95)
    yes
    log2x+log4x=2

    so log2x+log4x

    is the same as log2x times log4x

    so its log8x=2
  8. avtar95's Avatar
    8 15-04-2012  02:00
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    Re: Logs
    (Original post by dongonaeatu)
    log2x+log4x=2

    so log2x+log4x

    is the same as log2x times log4x

    so its log8x=2
    yes
  9. avtar95's Avatar
    9 15-04-2012  02:01
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    • Posts: 146
    Re: Logs
    (Original post by dongonaeatu)
    log2x+log4x=2

    so log2x+log4x

    is the same as log2x times log4x

    so its log8x=2
    would it be 64
  10. dongonaeatu's Avatar
    10 15-04-2012  02:03
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    Re: Logs
    (Original post by avtar95)
    would it be 64
    yes
  11. avtar95's Avatar
    11 15-04-2012  02:07
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    Re: Logs
    (Original post by dongonaeatu)
    yes

    :confused: in the answer book it is 2.52
  12. dongonaeatu's Avatar
    12 15-04-2012  02:07
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    Re: Logs
    (Original post by avtar95)
    :confused: in the answer book it is 2.52
    oh... sorry im not that good with logs

    its definitely log8x though
  13. Joshmeid's Avatar
    13 15-04-2012  02:09
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    Re: Logs
    log2x + log4x = 2

    log2x + log2x/log24 = 2

    log2x + log2x/2 = 2

    2log2x + log2x = 4

    3log2x = 4

    log2x^3 = 4

    x^3 = 16

    x = 2.52(3sf)

    Ask if you are confused by anything.
    Post rating:
    1
  14. avtar95's Avatar
    14 15-04-2012  02:10
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    • Posts: 146
    Re: Logs
    (Original post by dongonaeatu)
    oh... sorry im not that good with logs

    its definitely log8x though
    it's on the c2 cd
  15. dongonaeatu's Avatar
    15 15-04-2012  02:14
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    Re: Logs
    (Original post by Joshmeid)
    log2x + log4x = 2

    log2x + log2x/log24 = 2

    log2x + log2x/2 = 2

    2log2x + log2x = 4

    3log2x = 4

    log2x^3 = 4

    x^3 = 16

    x = 2.52(3sf)

    Ask if you are confused by anything.
    dont understand the second step
  16. avtar95's Avatar
    16 15-04-2012  02:16
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    • Posts: 146
    Re: Logs
    (Original post by Joshmeid)
    log2x + log4x = 2

    log2x + log2x/log24 = 2

    log2x + log2x/2 = 2

    2log2x + log2x = 4

    3log2x = 4

    log2x^3 = 4

    x^3 = 16

    x = 2.52(3sf)

    Ask if you are confused by anything.
    y did u do log2x / log24
  17. raheem94's Avatar
    17 15-04-2012  02:16
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    • Posts: 5,512
    Re: Logs
    (Original post by dongonaeatu)
    log2x+log4x=2

    so log2x+log4x

    is the same as log2x times log4x

    so its log8x=2
    No, you are making mistake again.

    log_2x+log_4x \not= log_2x \times log_4x
  18. raheem94's Avatar
    18 15-04-2012  02:17
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    Re: Logs
    (Original post by avtar95)
    y did u do log2x / log24
    The change of base rule.

    \displaystyle log_4x = \frac{log_2x}{log_24}
  19. dongonaeatu's Avatar
    19 15-04-2012  02:18
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    Re: Logs
    (Original post by raheem94)
    No, you are making mistake again.

    log_2x+log_4x \not= log_2x \times log_4x
    is that rule just for log2x and log4x i've never been taught that
  20. avtar95's Avatar
    20 15-04-2012  02:19
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    • Posts: 146
    Re: Logs
    (Original post by raheem94)
    The change of base rule.

    \displaystyle log_4x = \frac{log_2x}{log_24}
    could u put that in letters
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