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Math Help - Solving Logs

  1. #1
    Junior Member
    Joined
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    Solving Logs

    Find the solution of the logarithmic equation
    log x + log (x-2) = log (2x)

    In terms of logarithms, or correct to four decimal places.
    x= _____________



    Solve for x:
    log (x^5) = (log x)^2

    Note there are two solutions, A and B, where A < B
    A= ________
    B =________



    Rewrite the expression:
    4 log x - 2 log (x^2 +1) + 5 log (x-1)

    as a single logarithm log A. then function
    A = _________
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  2. #2
    Super Member redsoxfan325's Avatar
    Joined
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    Swampscott, MA
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    Quote Originally Posted by B-lap View Post
    Find the solution of the logarithmic equation
    log x + log (x-2) = log (2x)

    In terms of logarithms, or correct to four decimal places.
    x= _____________



    Solve for x:
    log (x^5) = (log x)^2

    Note there are two solutions, A and B, where A < B
    A= ________
    B =________



    Rewrite the expression:
    4 log x - 2 log (x^2 +1) + 5 log (x-1)

    as a single logarithm log A. then function
    A = _________
    You know the rule \log a+\log b=\log(ab)?

    So \log x +\log(x-2)=\log(2x)\implies\log(x^2-2x)=\log(2x) \implies x^2-2x=2x\implies x(x-4)=0\implies x=\{0,4\}. But x=0 doesn't work, because in the original equation, that would make the first term \log0, which is undefined. So x=4 is the only solution.

    Using also the rules that \log a-\log b=\log\frac{a}{b} and \log(a^b)=b\log a, see if you can do the others.
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