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Simplify $ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    As in title, simplify [tex]ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)[/tex], x > 0.


    2. Relevant equations
    -


    3. The attempt at a solution
    So far
    [tex]ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1) = ln(x(\sqrt{1+e^x}-\sqrt{e^x})(\sqrt{1+e^{-x}}+1)) = ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x}))[/tex].

    Is it possible to get further?
     
  2. jcsd
  3. Feb 22, 2009 #2

    Mark44

    Staff: Mentor

    I worked it through and ended with what you have. I don't see anything more you can do.
     
  4. Feb 22, 2009 #3
    Thanks. :)
     
  5. May 8, 2009 #4
    Actually, it's possible to get further:

    [itex]\ln \left(x \left(\sqrt{1+e^{x}}-\sqrt{e^{x}}\right) + \ln\left(\sqrt{1+e^{-x}}+1 \right) =
    \ln x + \ln\left(\sqrt{1+e^{x}}-\sqrt{e^{x}}\right) + \ln\left(\sqrt{1+e^{-x}}+1\right) =[/itex]

    [itex]= \ln x + \ln\left(\left(\sqrt{1+e^{x}}-\sqrt{e^x}\right) \left(\sqrt{1+e^{-x}} + 1\right)\right)
    = \ln x + \ln\left(\sqrt{1+e^{x}}\sqrt{1+e^{-x}}+\sqrt{1+e^{x}}-\sqrt{e^{x}}\sqrt{1+e^{-x}}-\sqrt{e^x}\right)[/itex]

    [itex]= \ln x + \ln\left(\left(\left(1+e^{x}\right)\left(1+e^{-x}\right)\right)^{\frac{1}{2}} + \left(1+e^{x}\right)^{\frac{1}{2}}-\left(\left(e^{x}\right)\left(1+e^{-x}\right)\right)^{\frac{1}{2}}-\left(e^{x}\right)^{\frac{1}{2}}\right) = \ln x + \ln\left(\left(1+e^{-x}+e^{x}+e^{0}\right)^{\frac{1}{2}} + \left(1+e^{x}\right)^{\frac{1}{2}} - \left(e^{x}+e^{0}\right)^{\frac{1}{2}}-\left(e^{x}\right)^{\frac{1}{2}}\right)[/itex]

    [itex]= \ln x + \ln\left(\left(2+e^{-x}+e^{x}\right)^{\frac{1}{2}} + \left(1+e^{x}\right)^{\frac{1}{2}} - \left(e^{x}+1\right)^{\frac{1}{2}} -\left(e^{x}\right)^{\frac{1}{2}}\right)
    = \ln x + \ln\left(\sqrt{\left(2 + e^{-x} + e^{x}\right)} - \sqrt{\left(e^{x}\right)}\right)[/itex]

    [itex]= \ln x + \ln\left(\sqrt{\left(e^{\frac{x}{2}} + e^{-\frac{x}{2}}\right)^2} - \sqrt{e^x}\right) = \ln x + \ln\left(e^{\frac{x}{2}} + e^{-\frac{x}{2}} - e^\frac{x}{2}\right)
    = \ln x + \ln\left(e^{-\frac{x}{2}}\right) = \ln \left(\frac{x}{e^\frac{x}{2}}\right)[/itex]
     
  6. May 8, 2009 #5

    Mark44

    Staff: Mentor

    You're right. It's also possible to get to your result with much less work.
    Here's where the OP ended.
    [tex] ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x}))[/tex]
    The expression in the first radical happens to be a perfect square trinomial, something I didn't notice earlier.


    [tex]2+e^{-x} + e^{x} = e^{x} + 2 + e^{-x} = (e^{x/2} + e^{-x/2})^2[/tex]
    The expression being squared is always positive, so we don't have to worry about the negative square root. IOW,
    [tex]\sqrt{e^{x} + 2+e^{-x} } = \sqrt{(e^{x/2} + e^{-x/2})^2} = e^{x/2} + e^{-x/2}[/tex]

    Here's the complete work.
    [tex]ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x}))
    = ln(x(e^{x/2} + e^{-x/2} - e^{x/2}))
    = ln(xe^{-x/2})
    = ln(\frac{x}{e^{x/2}})[/tex]
     
    Last edited: May 8, 2009
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