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    Hi Guys

    Its been along time since l've needed to post on the student forum, but people have always been helpful in the past so here...

    I've had an absolutely crap math lecturer at Uni this year so much so he's just been sacked and l have exams pending please can someone literally show me how to do this simplifying log

    log19 x4 y-2 -log19 x-1 y3 +log4 16

    I would be very grateful l need a step by step guide on this please as maths isn't my subject area but a hoop l must jump through in first year

    Many thanks in advance

    Karen
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    (Original post by Karen72)
    Hi Guys

    Its been along time since l've needed to post on the student forum, but people have always been helpful in the past so here...

    I've had an absolutely crap math lecturer at Uni this year so much so he's just been sacked and l have exams pending please can someone literally show me how to do this simplifying log

    log19 x4 y-2 -log19 x-1 y3 +log4 16

    I would be very grateful l need a step by step guide on this please as maths isn't my subject area but a hoop l must jump through in first year

    Many thanks in advance

    Karen
    \log_a b - \log_a c = \log_ a (b/c)

    and similarly,

    \log_ q m + \log_q n  = \log_q (mn)
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    i love how all the "URGENT HELP NOW" are always maths
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    (Original post by Indeterminate)
    \log_a b - \log_a c = \log_ a (b/c)

    and similarly,

    \log_ q m + \log_q n  = \log_q (mn)
    Really sorry but that means nothing to me at all, l am not a maths person on any level, l need someone to show me exactly what to do
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    (Original post by Karen72)
    Really sorry but that means nothing to me at all, l am not a maths person on any level, l need someone to show me exactly what to do
    Okay, I'll give you an example.

    \log_{19} 7 - \log_{19} 12 = \log_{19} \frac{7}{12}

    \log_4 (x+4) + \log_4 (x+2) = \log_4 [(x+4) \times (x+2)]

    Note that the base (number at the bottom right of the 'log') of two logs has to be the same for you to associate them in this way. It doesn't matter if you have symbols or numbers in the logs, as I've demonstrated to you above.

    Shall I go into some more detail or is that ok?
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    (Original post by Indeterminate)
    Okay, I'll give you an example.

    \log_{19} 7 - \log_{19} 12 = \log_{19} \frac{7}{12}

    \log_4 (x+4) + \log_4 (x+2) = \log_4 [(x+4) \times (x+2)]

    Note that the base (number at the bottom right of the 'log') of two logs has to be the same for you to associate them in this way. It doesn't matter if you have symbols or numbers in the logs, as I've demonstrated to you above.

    Shall I go into some more detail or is that ok?
    Thanks for your help! l get it up to there but am not sure what to do with the log4 16 do l need to simplify that as well?
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    (Original post by hexagon999)
    Hahahahahah.... I could do this maths when I was in prep school!
    seems unlikely . . .
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    (Original post by Karen72)
    Thanks for your help! l get it up to there but am not sure what to do with the log4 16 do l need to simplify that as well?
    \log_4 16 = 2

    This is because 4^2=16

    If a^x =k then \log_a k = x
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    (Original post by Karen72)
    Thanks for your help! l get it up to there but am not sure what to do with the log4 16 do l need to simplify that as well?
    Yes. Let me give you an example again

    \log_2 8 = 3

    because

    2^3 = 8 = 2 \times 2 \times 2

    You have to find out how many times the base has to be multiplied by itself to give the number that is being logged (in your case, this is 16).
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    (Original post by Mr M)
    \log_4 16 = 2

    This is because 4^2=16

    If a^x =k then \log_a k = x
    Thank for all the helpful posts from different people, very much appreciate it

    Karen
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    (Original post by Indeterminate)
    Yes. Let me give you an example again

    \log_2 8 = 3

    because

    2^3 = 8 = 2 \times 2 \times 2

    You have to find out how many times the base has to be multiplied by itself to give the number that is being logged (in your case, this is 16).
    Thanks very much you've been a star!
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    (Original post by hexagon999)
    I have an IQ of over 600, so...
    I reiterate - seems unlikely.
 
 
 
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