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    help i cant answer this at all
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    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.
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    (Original post by Tishax2)
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    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)
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    (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 ?
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    (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
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    (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.
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    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
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    (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}
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    (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
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    (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.
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    (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
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    (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.
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    (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
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    (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
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    (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
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    (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.
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    (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.
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    (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}
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    (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
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    (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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