C2 edexcel log question

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  1. Tishax2's Avatar
    1 02-05-2012  16:50
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    C2 edexcel log question
    Click image for larger version.  Name: problem4.png  Views: 126  Size: 1.1 KB  ID: 145476

    help i cant answer this at all
  2. notnek's Avatar
    2 02-05-2012  16:55
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    Re: C2 edexcel log question
    What is 3 written as a base 2 logarithm? Then combine the logs on the right hand side and rewrite the log on the left hand side. You should end up with something like:

    \log_2 A = \log_2 B

    And then you can say that A=B.
  3. raheem94's Avatar
    3 02-05-2012  16:56
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    Re: C2 edexcel log question
    (Original post by Tishax2)
    Click image for larger version.  Name: problem4.png  Views: 126  Size: 1.1 KB  ID: 145476

    help i cant answer this at all
    \displaystyle 2log_2y=3+log_2(y+6) \implies log_2y^2 - log_2(y+6) =3

    Now use the rule, \displaystyle log_ab-log_ac = log_a\left(\frac{b}{c}\right)
  4. Tishax2's Avatar
    4 02-05-2012  17:09
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    Re: C2 edexcel log question
    (Original post by raheem94)
    \displaystyle 2log_2y=3+log_2(y+6) \implies log_2y^2 - log_2(y+6) =3

    Now use the rule, \displaystyle log_ab-log_ac = log_a\left(\frac{b}{c}\right)
    what do i do from there ?
  5. Tishax2's Avatar
    5 02-05-2012  17:10
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    Re: C2 edexcel log question
    (Original post by notnek)
    What is 3 written as a base 2 logarithm? Then combine the logs on the right hand side and rewrite the log on the left hand side. You should end up with something like:

    \log_2 A = \log_2 B

    And then you can say that A=B.


    3 is jsut a number

    and y >0
  6. notnek's Avatar
    6 02-05-2012  17:15
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    • Location: Bangkok, Thailand
    Re: C2 edexcel log question
    (Original post by Tishax2)
    3 is jsut a number

    and y >0
    3 is just a number and so is log_2 2^3 which is equal to 3 (y>0 doesn't change this).

    So you have:

    \displaystyle 2\log_2 y = \log_2 8 + \log_2 (y+6)

    Can you follow the next steps in my last post?

    Also, it's a good idea for you to try both methods that have been given to you in this thread so you can decide which one you are more comfortable with.
    Last edited by notnek; 02-05-2012 at 17:17.
  7. aeyurttaser13's Avatar
    7 02-05-2012  17:17
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    Re: C2 edexcel log question
    2log2 y - log2 y+6 =3
    2log2 y/(y+6) =3
    log 2 (y/(y+6)squared = 3
    2 cubed = (y/(y+6)) squared.. the rest u should be able to work out
  8. notnek's Avatar
    8 02-05-2012  17:20
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    Re: C2 edexcel log question
    (Original post by aeyurttaser13)
    2log2 y - log2 y+6 =3
    2log2 y/(y+6) =3
    log 2 (y/(y+6)squared = 3
    2 cubed = (y/(y+6)) squared.. the rest u should be able to work out
    You've made a mistake in your second line although it's hard to know for sure because your working isn't clear.

    You cannot combine two logs which have different coefficients. In general,

    \displaystyle 2\log_c a - \log_c b \neq 2\log_c \frac{a}{b}
    Last edited by notnek; 02-05-2012 at 17:24.
  9. Tishax2's Avatar
    9 02-05-2012  17:23
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    Re: C2 edexcel log question
    (Original post by notnek)
    3 is just a number and so is log_2 2^3 which is equal to 3 (y>0 doesn't change this).

    So you have:

    \displaystyle 2\log_2 y = \log_2 8 + \log_2 (y+6)

    Can you follow the next steps in my last post?

    Also, it's a good idea for you to try both methods that have been given to you in this thread so you can decide which one you are more comfortable with.
    im sorry i really do not understand your method

    and especically why you put 3 as a power when it says 3+...

    thanks
  10. raheem94's Avatar
    10 02-05-2012  17:25
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    Re: C2 edexcel log question
    (Original post by Tishax2)
    what do i do from there ?
    \displaystyle 2log_2y=3+log_2(y+6) \implies log_2y^2 - log_2(y+6) =3 \implies log_2\left(\frac{y^2}{y+6}\right )=3

    Now use the rule, log_ab=x \implies a^x = b

    You will get a quadratic so you will obtain 2 solutions, reject the value which is less than zero.
  11. raheem94's Avatar
    11 02-05-2012  17:27
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    Re: C2 edexcel log question
    (Original post by Tishax2)
    im sorry i really do not understand your method

    and especically why you put 3 as a power when it says 3+...

    thanks
    Remember, 3 = log_22^3 = 3log_22
  12. notnek's Avatar
    12 02-05-2012  17:29
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    • Location: Bangkok, Thailand
    Re: C2 edexcel log question
    (Original post by Tishax2)
    im sorry i really do not understand your method

    and especically why you put 3 as a power when it says 3+...

    thanks
    No problem. I'll try to explain it further.

    What I have done is say that 3 is equal to log_2 8. In your first lessons on logarithms, you should have learnt why this is true. Put log_2 8 into your calculator to convince yourself if you're still not sure.

    Remember, log_2 2^1 = 1, log_2 2^2 = 2, log_2 2^3 = 3 etc.

    Next I have just replaced 3 with log_2 8, since they are the same thing.
  13. aeyurttaser13's Avatar
    13 02-05-2012  17:31
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    • Location: Bahrain
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    Re: C2 edexcel log question
    (Original post by notnek)
    You've made a mistake in your second line although it's hard to know for sure because your working isn't clear.

    You cannot combine two logs which have different coefficients. In general,

    \displaystyle 2\log_c a - \log_c b \neq 2\log_c \frac{a}{b}
    oops, ib math exam tmrw and still stupid mistakes! ughh thx for pointing it out its like a warning before my exam tmrw that i gotta DOUBLE CHECK haha
    shouldve been log_c a squared - log_c b = log_c a (squared) / b
  14. Tishax2's Avatar
    14 02-05-2012  17:41
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    Re: C2 edexcel log question
    (Original post by raheem94)
    \displaystyle 2log_2y=3+log_2(y+6) \implies log_2y^2 - log_2(y+6) =3 \implies log_2\left(\frac{y^2}{y+6}\right )=3

    Now use the rule, log_ab=x \implies a^x = b

    You will get a quadratic so you will obtain 2 solutions, reject the value which is less than zero.
    how do i get the quad equation from here :s
  15. Tishax2's Avatar
    15 02-05-2012  17:41
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    • Posts: 104
    Re: C2 edexcel log question
    (Original post by notnek)
    No problem. I'll try to explain it further.

    What I have done is say that 3 is equal to log_2 8. In your first lessons on logarithms, you should have learnt why this is true. Put log_2 8 into your calculator to convince yourself if you're still not sure.

    Remember, log_2 2^1 = 1, log_2 2^2 = 2, log_2 2^3 = 3 etc.

    Next I have just replaced 3 with log_2 8, since they are the same thing.
    thanks but we did not learn that in class :s
  16. Jay™'s Avatar
    16 02-05-2012  17:47
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    • Location: Birmingham
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    Re: C2 edexcel log question
    (Original post by Tishax2)
    how do i get the quad equation from here :s
    y^2 / (y+6) = 2^3

    y^2 / (y+6) = 8

    y^2 = 8(y+6)

    y^2 = 8y + 48

    y^2 - 8y - 48 = 0

    And solve.
  17. notnek's Avatar
    17 02-05-2012  17:53
    • TSR Demigod
    • Location: Bangkok, Thailand
    Re: C2 edexcel log question
    (Original post by Tishax2)
    how do i get the quad equation from here :s
    \displaystyle \log_2 \frac{y^2}{y+6} = 3 \implies \frac{y^2}{y+6}=2^3

    If this confuses you then what's happening is similar to e.g this:

    \displaystyle \log_2 x = 5 \implies x=2^5

    (5 is an arbitrary number that I picked out of the air to show you an example)

    Do you understand this? If not, you may need to revise some basic logarithms work.
  18. raheem94's Avatar
    18 02-05-2012  17:53
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    Re: C2 edexcel log question
    (Original post by Tishax2)
    how do i get the quad equation from here :s
    \displaystyle log_2\left(\frac{y^2}{y+6}\right )=3 \implies 2^3=\frac{y^2}{y+6}
  19. Tishax2's Avatar
    19 02-05-2012  17:57
    • Full Member
    • Posts: 104
    Re: C2 edexcel log question
    (Original post by raheem94)
    \displaystyle 2log_2y=3+log_2(y+6) \implies log_2y^2 - log_2(y+6) =3 \implies log_2\left(\frac{y^2}{y+6}\right )=3

    Now use the rule, log_ab=x \implies a^x = b

    You will get a quadratic so you will obtain 2 solutions, reject the value which is less than zero.
    i did it and i got y=12
  20. Tishax2's Avatar
    20 02-05-2012  17:57
    • Full Member
    • Posts: 104
    Re: C2 edexcel log question
    (Original post by raheem94)
    \displaystyle log_2\left(\frac{y^2}{y+6}\right )=3 \implies 2^3=\frac{y^2}{y+6}
    y=12
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