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Math Help - Logs

  1. #1
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    Logs

    Solve for x:
    2log(x)-log(x=1)= log(4)-log(3)
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  2. #2
    Senior Member
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    Hi Soulwraith56

    Do you mean : 2*log(x)-log(x+1)= log(4)-log(3) ?
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  3. #3
    Junior Member
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    OMG, I love solving logs!!

    Are you familiar with the log laws Soulwraith? If not, I strongly advise you thoroughly revise them!

    As for your question:
    2*log(x)-log(x+1)= log(4)-log(3)

    Here is a process to work it out (Assuming log is to the base 10):
    2*\log x-\log (x+1)=\log 4-\log 3

    Simplifying the logs into single terms using log laws:
    log x^2-\log (x+1)=\log (4/3)
    log (x^2/x+1)=\log (4/3)

    Taking the log of both sides leaves us with an equation we can solve for x:

    (x^2)/(x+1)=\frac{4}{3}

    Can you take it from here?
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  4. #4
    Senior Member pacman's Avatar
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    2log(x)-log(x+1)= log(4)-log(3), if this correct, we may continue

    rule of logarithm

    i) log xy = log x + log y
    ii) log x/y = log x - log y
    iii) log x^n = n log x

    2log(x)-log(x+1)= log(4)-log(3),

    use rule iii)

    log x^2 - log (x + 1) = log 4 - log 3

    use rule ii)

    log ((x^2)/(x + 1) = log 4/3

    take anti-log of both sides

    (x^2)/(x + 1) = 4/3

    cross-multiply

    3x^2 = 4(x + 1),

    3x^2 = 4x + 4,

    3x^2 - 4x - 4 = 0,

    factoring, we have

    (x - 2)(3x + 2) = 0

    x = {-2/3, 2}

    but only x = 2 is valid.

    for x to have the value of -2/3 is multivalued in logarithm.

    the graph of log x^2 - log (x + 1) = log 4 - log 3 is shown below.

    It confirms that x = 2 is only the valid value of x.
    Attached Thumbnails Attached Thumbnails Logs-log.gif  
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