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Math Help - [GCSE-O] Logarithms And Indices

  1. #1
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    [GCSE-O] Logarithms And Indices

    1)
    Given y2 - 1 = x2
    Show that log(y-1)x + log(y+1)x = 2(log(y-1)x)(log(y+1)x)
    Here are my steps
    Please let me know if I make any mistakes

    log(y-1)(y+1)[(y-1)(y+1)]= log(y-1)(y+1)x2
    log(y-1)(y+1)[(y-1)(y+1)]= 2 log(y-1)(y+1)x

    log(y-1)[(y-1)(y+1)] / log(y+1)[(y-1)(y+1)] = 2 ( log(y-1)(y+1)x )

    log(y-1)(y+1) / log(y+1)(y-1) = 2 ( log(y-1)(y+1)x )

    I got stuck here. I am not sure how to move forward from here.

    2)
    Find the value of x such that (2/x)logn4 - (3/x)logn9 = 0
    x2 - 2y2 = 2x - 3y
    Here are my steps

    logn4 [ logn(2/x) ] = logn9 [ logn (3/x) ]
    logn22 [ logn(2/x) ] = logn32 [ logn (3/x) ]
    logn22 [ logn(2) - logn(x)] = logn32 [ logn(3) - logn(x) ]
    logn22 logn(2) - logn22 logn(x) = logn32 logn(3)- logn32 logn(x)

    I got stuck here and I am not sure how to move forward from here.

    3)
    If 2x3y = 3x4y = 6
    show by taking logarithms to base 2 and eliminating log23, that
    x2 - 2y2 = 2x - 3y
    Here are my following steps
    2x3y = 3x4y = 6
    2x/3x = 4y/3y
    (2/3)x = (4/3)y

    x log2(2/3) = y log2(4/3)
    x [ log2(2) - log2(3) ] = y [ log2(4) - log2(3) ]
    x log2(2) - 2y log2(2) = x log2(3) - y log2(3)
    x - 2y = log2(3) [(x-y)]
    log2(3) = (x - y)/(x - 2y)

    I know I did something wrong here but I just couldn't figure out.

    Thank you very much. Hope someone can help me out! I will appreciate it. I don't need a full solution but I would like some hints on how should I carry on. Thank you!!
    Last edited by mynameiscosine; June 20th 2013 at 10:31 AM.
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  2. #2
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    Re: [GCSE-O] Logarithms And Indices

    Hello, mynameiscosine!

    Here is #3.


    \text{3. Given: }\:2^x3^y \:=\:3^x4^y \:=\:6

    \text{show by taking logs (base 2) and eliminating }\log_23

    . . \text{that: }\:x^2-2y^2 \:=\:2x - 3y

    We have: . 2^x3^y \:=\:6

    Take logs, base 2: . \log_2(2^x3^y) \:=\:\log_2(6)

    n . . . . . . . \log_2(2^x) + \log_2(3^y) \:=\:\log_2(2\cdot3)

    n . . . . . . . . x\underbrace{\log_22}_1 + y\log_23 \:=\:\underbrace{\log_22}_1 + \log_23

    . . . . . . . . . . . . . x + y\log_23 \:=\:1 + \log_23

    . . . . . . . . . . y\log_23 - \log_23 \:=\:1 - x

    . . . . . . . . . . . . (y-1)\log_23 \:=\:1-x

    . . . . . . . . . . . . . . . . . \log_23 \:=\:\frac{1-x}{y-1} .[1]


    We have: . 3^x(2^2)^y \:=\:6 \quad\Rightarrow\quad 3^x2^{2y} \:=\:6

    Take logs, base 2: . \log_2(3^x2^{2y}) \:=\:\log_2(6)

    n . . . . . . . \log_2(3^x) + \log_2(2^{2y}) \:=\:\log_2(2\cdot 3)

    . . . . . . . . . x\log_23 + 2y\underbrace{\log_22}_1 \:=\:\underbrace{\log_22}_1 + \log_23

    . . . . . . . . . . . x\log_23 - \log_23 \:=\:1 - 2y

    n . . . . . . . . . . . (x-1)\log_23 \:=\:1-2y

    . . . . . . . . . . . . . . . . . \log_23 \:=\:\frac{1-2y}{x-1} .[2]


    Equate [1] and [2]: . \frac{1-x}{y-1} \:=\:\frac{1-2y}{x-1}

    . . . . . . . . . (1-x)(x-1) \:=\:(y-1)(1-2y)

    . . . . . . . . . x - 1 - x^2 + x \:=\: y - 2y^2 - 1 + 2y

    . . . . . . . . . . . . . x^2 - 2y^2 \:=\: 2x - 3y
    Thanks from mynameiscosine
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  3. #3
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    Re: [GCSE-O] Logarithms And Indices

    @Soroban

    Thank you so much!
    I am still working on the other 2 questions.

    Thanks! (:

    By the way, how do I do the fractions symbol? I can't seems to find it. Thanks!
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