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Math Help - Help with logarithms

  1. #1
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    Help with logarithms

    I am really stumped on this question. I have tried re arranging it but I can't seem to get it right.

    ln x - ln x^2 = ln 27
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  2. #2
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    Re: Help with logarithms

    logarithms
    ln a - ln b= ln a/b
    ln 1= 0
    ln a +ln b= ln ab
    ln e= 1
    ln a= ln b is a=b
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  3. #3
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    Re: Help with logarithms

    logarithms
    ln a - ln b= ln a/b
    ln 1= 0
    ln a +ln b= ln ab
    ln e= 1
    ln a= ln b is a=b
    Follow Math Help Forum on Facebook and Google+

  4. #4
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    Re: Help with logarithms

    sorry for the double on posts.
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  5. #5
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    Re: Help with logarithms

    Not much help. I know the rules. I just can't solve it. Running out of time.
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  6. #6
    MHF Contributor Reckoner's Avatar
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    Re: Help with logarithms

    Quote Originally Posted by SplashDamage View Post
    Not much help. I know the rules. I just can't solve it. Running out of time.
    At what point do you get stuck? Are you able to combine the left-hand side into a single logarithm? After that you can equate the expressions inside the logs (or, equivalently, make both sides exponential to the base e so that the logarithm operations get canceled out).
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  7. #7
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    Re: Help with logarithms

    I'm interested in knowing how to solve this one too.
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  8. #8
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    Re: Help with logarithms

    Quote Originally Posted by SplashDamage View Post
    I am really stumped on this question. I have tried re arranging it but I can't seem to get it right.

    ln x - ln x^2 = ln 27
    With the rule \ln(a)-\ln(b)=\ln\left(\frac{a}{b}\right) the equation is equivalent to:
    \ln\left(\frac{x}{x^2}\right)=\ln(27)
    \Leftrightarrow \ln\left(\frac{1}{x}\right)=\ln(27)
    Therefore \frac{1}{x}=27 and thus x=\frac{1}{27} is the solution.
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