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Math Help - Coverting to a single logarithm

  1. #1
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    Question Coverting to a single logarithm

    i begin with this question
    Express, log(x+1) - log(x-6) + 2 log x
    as a single logarithm
    so far i have come to an answer which is slightly puzzeling but i somehow mixed the expression to become equal to something which could be why i had the wrong answer, i have attempted this again, but im at a stage where im getting stuck, any chance for someone to correct me =]

    workings
    log(x+1/x-6) +logx^2
    so now i have to times the two logs together
    logx^2(x+1/x-6)
    logx^2(x+1) divided by logX^2(x-6)......
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  2. #2
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    Quote Originally Posted by JorCherryhead View Post
    workings
    log(x+1/x-6) +logx^2
    so now i have to times the two logs together
    logx^2(x+1/x-6)
    You are right up to this point

    \log (x+1) - \log (x-6) + 2 \log x = \log \left(\frac{x+1}{x-6}\right)  + \log x^2 = \log \left(\frac{x^2(x+1)}{x-6}\right)
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  3. #3
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    Quote Originally Posted by JorCherryhead View Post
    i begin with this question
    Express, log(x+1) - log(x-6) + 2 log x
    as a single logarithm
    so far i have come to an answer which is slightly puzzeling but i somehow mixed the expression to become equal to something which could be why i had the wrong answer, i have attempted this again, but im at a stage where im getting stuck, any chance for someone to correct me =]

    workings
    log(x+1/x-6) +logx^2
    so now i have to times the two logs together
    logx^2(x+1/x-6)
    logx^2(x+1) divided by logX^2(x-6)......
    You're adding the log(x^2) so you should have it in the numerator

    \log \left(\frac{x^2(x+1)}{x-6}\right)
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