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    (b) Calculate the value of y for which 2 log3 y − log3 ( y + 4 ) = 2 .
    Here's my working out:

    2 log3 y - log3 y - log3 4 = 2

    log3 y - log3 4 = 2

    log3 (y/4)=2

    3^2=y/4

    9=y/4

    y=36

    The correct answer is y=12. WHY isn't the above correct?? What have I done wrong?

    THANKS
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    (Original post by krisshP)
    (b) Calculate the value of y for which 2 log3 y − log3 ( y + 4 ) = 2 .
    Here's my working out:

    2 log3 y - log3 y - log3 4 = 2

    log3 y - log3 4 = 2

    log3 (y/4)=2

    3^2=y/4

    9=y/4

    y=36

    The correct answer is y=12. WHY isn't the above correct?? What have I done wrong?

    THANKS
    log_{3} (y+4) \not= log_{3}y + log_{3}4

    As  log_{a} (xy) = log_{a}x + log_{a}y.

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    (Original post by krisshP)
    (b) Calculate the value of y for which 2 log3 y − log3 ( y + 4 ) = 2 .
    Here's my working out:

    2 log3 y - log3 y - log3 4 = 2

    log3 y - log3 4 = 2

    log3 (y/4)=2

    3^2=y/4

    9=y/4

    y=36

    The correct answer is y=12. WHY isn't the above correct?? What have I done wrong?

    THANKS
    The line indicated
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    Can you explain why please? I need to actually understand. Why can't you just expand the brackets?
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    (Original post by krisshP)
    Can you explain why please? I need to actually understand. Why can't you just expand the brackets?
    They are not equal

    What do you want to understand

    The log function will not work in that way
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    (Original post by krisshP)
    Can you explain why please? I need to actually understand. Why can't you just expand the brackets?
    As  log_{a} (xy) = log_{a}x + log_{a}y

    log_{a}(x + y) \not= log_{a}x + log_{a}y


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    Okay, thanks
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    Interestingly, in the case a=3, there is a value of m and a value of n such that

    \log_3 m + \log_3 n = \log_3 (m+n)

    \log_3 (m^2) = (\log_3 m)^2
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    (Original post by Indeterminate)
    ...
    Bad ... Indeterminate ... Bad
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    (Original post by TenOfThem)
    Bad ... Indeterminate ... Bad
    What? Both aren't generally true, yes, but what I've said is correct for a specific value of m and a specific value of n.
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    (Original post by Indeterminate)
    What? Both aren't generally true, yes, but what I've said is correct for a specific value of m and a specific value of n.
    KrisshP needs no confusion
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    (Original post by TenOfThem)
    KrisshP needs no confusion


    A bit too cheeky, maybe.
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    (Original post by Indeterminate)
    Interestingly, in the case a=3, there is a value of m and a value of n such that

    \log_3 m + \log_3 n = \log_3 (m+n)

    \log_3 (m^2) = (\log_3 m)^2
    If it's any consolation, I thought it was pretty cool too. Shhh, don't tell ToT
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    (Original post by Occams Chainsaw)
    Shhh, don't tell ToT
    I can read LoL
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    (Original post by krisshP)
    Can you explain why please? I need to actually understand. Why can't you just expand the brackets?
    It's because a logarithm is a function - in the same way (2+3)^2 is not the same as (2^2) + (3^2).
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    (Original post by TenOfThem)
    I can read LoL
    I figured you would only come back to the thread if you were quoted!
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    (Original post by Occams Chainsaw)
    I figured you would only come back to the thread if you were quoted!
    I can smell out my name even if it is abbreviated
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    Okay, so here is my working:

    2log3 y - log3(y+4) = 2
    log3 y^2 - log3(y+4) = 2
    log3( y^2 / y+4 ) = 2
    Take 3 to the power of both sides to give
    3^ (log3 y^2 / y+4 ) = 3^2
    y^2 / y+4 = 9
    multiply through by y + 4
    y^2 = 9 (y+4)
    y^2 = 9y = 36
    y^2 - 9y -36 = 0
    (y - 12) (y + 3) = 0
    so y = 12, -3
    can't do a negative number with logs, so y = 12 is the answer

    Hope this helped
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    (Original post by TenOfThem)
    I can smell out my name even if it is abbreviated
    I've got a good log question that my maths lectures couldn't/wouldn't answer for me so I'm wondering if you can/will please!

    Why do negative base values sometimes work but sometimes not? I presume it's got something to do with complex numbers but I can't make the connection tbh.

    Sooooo many things annoy me about A Level maths like this! Like why the CoM of a semi-circle is 4(r)/3(pi) and why exactly the way we integrating (namely dividing my the new power) actually works. Or how sine is actually derived (not that circle on an axis thing, that's cheating!).

    Mechanical use of the principles is stupid.. We need to know why, I think! Maybe I should be able to work it out for myself but I am far too lazy and probably not clever enough to do it!
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    (Original post by Occams Chainsaw)
    I've got a good log question that my maths lectures couldn't/wouldn't answer for me so I'm wondering if you can/will please!

    Why do negative base values sometimes work but sometimes not? I presume it's got something to do with complex numbers but I can't make the connection tbh.

    Sooooo many things annoy me about A Level maths like this! Like why the CoM of a semi-circle is 4(r)/3(pi) and why exactly the way we integrating (namely dividing my the new power) actually works. Or how sine is actually derived (not that circle on an axis thing, that's cheating!).

    Mechanical use of the principles is stupid.. We need to know why, I think! Maybe I should be able to work it out for myself but I am far too lazy and probably not clever enough to do it!
    Well, to answer your main Q

    \log_{-3} 9 = 2

    works but it's not of any use tbh
 
 
 
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