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Math Help - Write the Expression as a Logarithm of a Single Quantity

  1. #1
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    Smile Write the Expression as a Logarithm of a Single Quantity

    \frac{3}{2} ln(x^2+1)-ln(x+1)-ln(x-1)

    I know that ln\frac{a}{b}=lna-lnb

    But I don't know which part to divide first. Is there any way to know what part to divide first or is this somehow a trick question?

    Thanks so much!!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liz155 View Post
    \frac{3}{2} ln(x^2+1)-ln(x+1)-ln(x-1)

    I know that ln\frac{a}{b}=lna-lnb

    But I don't know which part to divide first. Is there any way to know what part to divide first or is this somehow a trick question?

    Thanks so much!!!
    it doesn't matter. you will get the same thing in the end (just mind the signs in front of the logs). also note that \ln a + \ln b = \ln ab and n \ln a = \ln \left(a^n \right)
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  3. #3
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    Sorry I meant to write this...

    \frac{3}{2} [ ln(x^2+1)-ln(x+1)-ln(x-1)]

    Do the parenthesis make a difference?
    So when I combine the stuff inside the parenthesis, I put the 3/2 as an exponent?

    ln[( x^2+1)(x-1)/(x+1)]^(3/2)

    Does that seem right?

    Thanks for your help!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liz155 View Post
    Sorry I meant to write this...

    \frac{3}{2} [ ln(x^2+1)-ln(x+1)-ln(x-1)]

    Do the parenthesis make a difference?
    So when I combine the stuff inside the parenthesis, I put the 3/2 as an exponent?

    ln[( x^2+1)(x-1)/(x+1)]^(3/2)

    Does that seem right?

    Thanks for your help!
    yes, the parentheses make a big difference. and you are incorrect. look at the rules and try gain. i told you to mind the signs in front of the logs, the middle log has a minus sign in front (you implicitly treated it as a plus sign)
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  5. #5
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    I see what I did wrong. Thank you!

    I got...

    ln(( x^2+1)/( x^2-1))^(3/2)

    Does this seem right?

    Thanks!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liz155 View Post
    I see what I did wrong. Thank you!

    I got...

    ln(( x^2+1)/( x^2-1))^(3/2)

    Does this seem right?

    Thanks!
    correct
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