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Math Help - combining logs

  1. #1
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    combining logs

    Express the given quantity as a single logarithm. ln(a + b) + ln(a - b) - 9 ln(c)

    Here is what I did:

    ln a + ln b + ln a - ln b - 9 ln c
    The ln b's cancel.

    ln a + ln a - 9 ln c
    Then combine to one logarithm.

    ln(a^2 - c^9)

    The outcome I got was incorrect. I was just wondering if you could point me in the right direction or point out my flaws. Thanks!
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  2. #2
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    Quote Originally Posted by ryan18 View Post
    Express the given quantity as a single logarithm. ln(a + b) + ln(a - b) - 9 ln(c)

    Here is what I did:

    ln a + ln b + ln a - ln b - 9 ln c
    The ln b's cancel.

    ln a + ln a - 9 ln c
    Then combine to one logarithm.

    ln(a^2 - c^9)

    The outcome I got was incorrect. I was just wondering if you could point me in the right direction or point out my flaws. Thanks!
    \ln(a+b) \ne \ln{a} + \ln{b}

    \ln(a \cdot b) = \ln{a} + \ln{b}

    I recommend that you review the properties of logarithms.
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  3. #3
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    WOW. I dont know how I convinced myself of that!

    Ok I got it now.

    ln((a+b)(a-b)) - ln c^9
    Factor out.

    ln(a^2 - b^2) - ln c^9
    Add in the c value.

    ln((a^2 - b^2)/c^9)

    sorry.
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  4. #4
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    Quote Originally Posted by ryan18 View Post
    Express the given quantity as a single logarithm. ln(a + b) + ln(a - b) - 9 ln(c)

    Here is what I did:

    ln a + ln b + ln a - ln b - 9 ln c
    The ln b's cancel.

    ln a + ln a - 9 ln c
    Then combine to one logarithm.

    ln(a^2 - c^9)

    The outcome I got was incorrect. I was just wondering if you could point me in the right direction or point out my flaws. Thanks!
    remember laws of logarithm...
    ln (a b)= ln a+ ln b
    ln  \frac{a}{b}= ln a- ln b
    so,
    ln a + ln b + ln a - ln b - 9 ln c= ln  \frac{a b a}{b c^9}
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