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Math Help - need help ASAP

  1. #1
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    need help ASAP

    find the exact value of x for which

    ln(x^2+4x+3)-ln(x^2+x)=4
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  2. #2
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    \ln (a) - \ln (b) = \ln \left(\frac{a}{b}\right)

    \ln (x^2+4x+3) - \ln (x^2+x) = 4
    \ln \left(\frac{x^{2} + 4x + 3}{x^{2} + x}\right) = 4

    Simplify the inside by factoring and canceling the common factor.

    This should help you right after: \ln (a) = b \iff \log_{e}a = b \iff a^{b} = e where e is the constant 2.718...

    Then it should be a matter of simple algebra to solve for x.
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  3. #3
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    Quote Originally Posted by outlaw View Post
    find the exact value of x for which

    ln(x^2+4x+3)-ln(x^2+x)=4

    ln(x^2+4x+3)-ln(x^2+x)=4

    ln((x+3)(x+1))-ln(x(x+1))=4

    ln(x+3) + ln(x+1)-(ln x + ln(x+1))=4

    ln(x+3) -ln x = 4

    ln((x+3)/x) = 4

    (x+3)/x = e^4

    x = 3/(e^4 - 1)
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