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    any one up for No.8 P2, jan. 2005. Need help! And need to get a grip with the whole Log thingy. Would appreciate it if any one could use this as a revision while Keying facts about 'Logs and it's extended Famlies' or outlining how to improve on them.
    Thanks
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    well post the question please if u didnt get the answer from anyone that has the question
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    (Original post by yazan_l)
    well post the question please if u didnt get the answer from anyone that has the question

    8(d). Show that the equation;

    1/2x) - 1 + ln(x/2)=0 can be rearranged into the form x = 2e ^(1 - (1/2x))

    2 marks question.
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    (1/2x) - 1 + ln(x/2) = 0
    ln(x/2) = 1- 1/2x

    you know that : loga b = c can be written as : ac = b
    so:ln(x/2) = 1- 1/2x
    loge[ x/2 ] = 1- 1/2x
    x/2 = e1- 1/2x
    x = 2 e1- 1/2x
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    'men' thanks a lot!
    U up 4 one more?
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    ok go one! dont ask, just post the question
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    ln sinx = -ln secx, in d interval 0<x<90
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    and this;

    log2 (2x+1) - log2x = 2
    find x,
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    ln sinx = -ln secx
    ln sinx = ln [secx]-1
    sinx = sec-1x
    sinx = cosx
    tanx = 1
    x = arctan1
    x = 45o
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    log2 (2x+1) - log2x = 2
    find x

    log2(2x+1) - log2x = 2

    logab - logac = loga(b/c)
    log2 [(2x+1)/x] = 2
    2² = [(2x+1)/x]
    4x = 2x + 1
    2x = 1
    x = ½
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    any body from JORDAN here????
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    above u!
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    thanks mate yazan, are u taking P2 also?
    and what tip do u think u can give in other 4 others & me to improve in that area(Logs)
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    no i've done P2
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    cool!
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    to be sure, lg - Log to d base 10, while ln is log to d base e, right?
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    ya right.
 
 
 
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Updated: June 18, 2005
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