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Math Help - Log Equation

  1. #1
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    Log Equation

    Okay, this is a (for you guys)simple Log Equation, the final is in a few day's and i've forgotten how to solve this. Forgotten some of the rules.



    I'm stuck here bty. - logx = log5 + log3 + log2
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  2. #2
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    Quote Originally Posted by butters View Post
    Okay, this is a (for you guys)simple Log Equation, the final is in a few day's and i've forgotten how to solve this. Forgotten some of the rules.



    I'm stuck here bty. - logx = log5 + log3 + log2
    \log{x} - \log(x-2) = \log{5} - \log(x+3)

    \log\left(\frac{x}{x-2}\right) = \log\left(\frac{5}{x-3}\right)<br />

    \frac{x}{x-2} = \frac{5}{x-3}

    finish up and solve for x ... don't forget to check your solution(s) in the original equation for domain issues.
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  3. #3
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    Hi there butters, consider the log law

     \ln(a)-\ln(b)= \ln\left(\frac{a}{b}\right)

    so

     \ln(x)-\ln(x-2)= \ln(5)-\ln(x+3)

     \ln\left(\frac{x}{x-2}\right)= \ln\left(\frac{5}{x+3}\right)

     \frac{x}{x-2}= \frac{5}{x+3}

    can you take it from here?
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  4. #4
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    Quote Originally Posted by skeeter View Post
    finish up and solve for x ... don't forget to check your solution(s) in the original equation for domain issues.
    I take it the initial steps i was taking in "logx = log5 + log3 + log2" were wrong..
    But i do remember solving problems like this in that method.
    Thank You.
    Quote Originally Posted by pickslides View Post

    can you take it from here?
    Yes, thank you, i'm getting "1 + or - 3i".
    The answer is given as "No solution". So does that mean, like negative number answers, the "i" make's it a "No solution" ?
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  5. #5
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    It means no real solution.
    Last edited by pickslides; December 15th 2009 at 07:48 PM. Reason: *sp
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  6. #6
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    Okay, once again, Thanks.
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