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Math Help - Solving equations

  1. #1
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    Solving equations

    I dont have any idea how to start this equation.



    ln(x^2-1) + 2ln(2) = ln (4x -1)


    How can i simplify this equation??


    (sorry my english)
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  2. #2
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    Re: Solving equations

    Quote Originally Posted by Fabio010 View Post
    I dont have any idea how to start this equation.



    ln(x^2-1) + 2ln(2) = ln (4x -1)


    How can i simplify this equation??


    (sorry my english)
    Are you familiar with the laws of logarithms?

    a\ln(k) = \ln(k^a)

    \ln(ab) = \ln(a) + \ln(b)

    and, because the logarithm is a one-to-one function: \ln(a) = \ln(b) \Leftrightarrow a = b
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  3. #3
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    Re: Solving equations

    never mind.


    I dont know why but in my head ln(a) + ln (b) = ln (ab) if a = b...

    sorry.
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  4. #4
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    Re: Solving equations

    What you do need to think about is that logarithm is a "one to one" function. That is, ln(a)= ln(b) if and only if a= b.
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  5. #5
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    Re: Solving equations

    Quote Originally Posted by Fabio010 View Post
    never mind.


    I dont know why but in my head ln(a) + ln (b) = ln (ab)

    sorry.
    That rule holds for all positive a and b. For example \ln(e) + \ln(e^3) = \ln(e \cdot e^3) = \ln(e^4) = 4 which is what we'd expect from adding them separately.
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  6. #6
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    Re: Solving equations

    So: ln(x^2-1) + 2ln(2) = ln (4x -1)

    ln(x^2-1) + ln(4) = ln (4x -1)
    ln ((x^2-1)4)) = ln(4x -1)
    ln ((4x^2-4)) = ln (4x-1)

    ln ((4x^2-4)) - ln (4x-1) = 0

    ln ((\frac{4x^2-4}{4x-1})) = 0
    \frac{4x^2-4}{4x-1} = e^0

    4x^2-4 = 4x-1 ?
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  7. #7
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    Re: Solving equations

    Quote Originally Posted by Fabio010 View Post
    So: ln(x^2-1) + 2ln(2) = ln (4x -1)

    ln(x^2-1) + ln(4) = ln (4x -1)
    ln ((x^2-1)4)) = ln(4x -1)
    ln ((4x^2-4)) = ln (4x-1)

    ln ((4x^2-4)) - ln (4x-1) = 0

    ln ((\frac{4x^2-4}{4x-1})) = 0
    \frac{4x^2-4}{4x-1} = e^0

    4x^2-4 = 4x-1 ?
    Yep, fine thus far. Now solve the quadratic bearing in mind any restrictions on the domain.
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  8. #8
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    Re: Solving equations

    Ok thanks for the help !
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  9. #9
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    Re: Solving equations

    Quote Originally Posted by Fabio010 View Post
    Ok thanks for the help !
    For reference the answer is x = \dfrac{3}{2}
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