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Math Help - Natural Log problem, please can you help?

  1. #1
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    Natural Log problem, please can you help?

    Hello
    I am new to this forum so therefore please accept my apologies if I have not posted this problem to the correct forum.

    x=5+ln(y/(y-1))


    I am trying to break out of the brackets to the above problem
    Please can someone confirm that I have gone about it correctly.
    Also can you confirm that I have solved for Y correctly. I am not confident with this at all.

    1) we know that natural log ln(a/b) = ln a - ln b therefore

    2) ln(y/(1-y)) = ln y - ln(1-y)

    = ln y - ln( -y + 1)

    = ln y - ln (1/y) + ln 1
    therefore

    x = ln y - ln 1 - ln y + ln 1 + 5


    3) Ok now to get Y on its own.

    ln y - ln 1 - ln y + ln 1 + 5 = x

    ln y - ln 1 + ln 1 - ln y + ln 1 + 5 = x + ln 1

    ln y - ln y + ln 1 + 5 = x + ln 1

    ln y - ln y + ln y + ln 1 + 5 = x + ln 1 + ln y

    ln y + ln 1 + 5 = x + ln 1 + ln y

    ln y + ln 1 - ln 1 + 5 = x + ln 1 + ln y - ln 1

    ln y + 5 - 5 = x + ln 1 + ln y - ln 1 - 5

    ln y = x + ln 1 + ln y - ln 1 - 5

    ln y / ln y = x + ln 1 + ln y - ln 1 - 5 / ln y

    so 1 or y = x + ln 1 + ln y - ln 1 - 5 / ln y

    This is how I am seeing it at the moment, I know that this cannot be right as the quotient equation for Y would be undefined as the denominator is equal to zero!

    Thanks for your time
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  2. #2
    A Plied Mathematician
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    = ln y - ln( -y + 1)

    = ln y - ln (1/y) + ln 1 therefore
    This step is incorrect. I'm not sure I would approach the problem this way. I'd go like this:

    x=5+ln(y/(y-1))

    x - 5 = ln(y/(y-1))

    e^(x-5) = y/(y-1)

    ...
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  3. #3
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    Thanks for your quick response
    I will give this way a try and let you know if I am still struggling.

    Thanks for your time
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  4. #4
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    Hi,
    Assuming you were correct when you wrote ln y - ln 1 - ln y + ln 1 + 5 = x, why didn't you say or think that ln y - ln y cancel out, ie equals 0 and that ln 1 - ln 1 also equals 0? So the left side is just 5 and so x = 5. You can't solve for Y since there are no y's since they canceled out.
    Your main problem is, as the previous poster stated so well, is that ln(-y + 1) can't be broken down. Yes ln (ab) = ln (a) + ln (b) and ln (a/b) = ln (a) - ln (b) but ln (a +/- b) is NOT ln (a) +/- ln (b). Now solve for ln(y/(y-1)) in x=5+ln(y/(y-1)).
    Good luck.
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