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Quick question about logarithims

  1. May 12, 2010 #1
    I solved for the equation

    [tex] \frac{f}{2} ln\frac{Tf}{Ti} = - ln \frac{vf}{vi} [/tex]

    I know to simplify the equation I need to Exponentiate both sides

    [tex]

    e^{\frac{f}{2}} * e^{ln\frac{Tf}{Ti}} = e^{ln\frac{vi}{vf}} [/tex]

    note i changed [tex] - ln \frac{vf}{vi} \textrm{ too } ln\frac{vi}{vf} [/tex] and removed the negative sign

    ,

    after exponentiating both sides I get

    [tex] e^{\frac{f}{2}} * \frac{Tf}{Ti} = \frac{Vi}{Vf} [/tex]

    But I'm not sure if I put the f/2 in the right place

    My equation should be reading
    [tex] {\frac{Tf}{Ti}}^\frac{f}{2} [/tex]
    but I don't know how to get to that
    from exponentiating [tex] \frac{f}{2} \frac{Tf}{Ti} [/tex]

    could someone please expain the rules behind this?

    --------------------------------------------

    original question:

    solve an equation for the work done in adiabatic compression

    -

    [tex] U = \frac{f}{2} N \kappa T [/tex]

    [tex] dU = \frac{f}{2} N \kappa dT [/tex]

    the work done during quasistatic compression is -P dv

    [tex] \frac{f}{2} N\kappa dT = -P dV [/tex]

    using the ideal gas law to write P the pressure in terms of the variables T and v
    ideal gas law: [tex] \frac{N \kappa T}{V} = P [/tex]

    i get

    [tex] \frac{f}{2} N\kappa dT = -\frac{N\kappa T}{V} dv [/tex]

    simplifying by canceling out N K gives

    [tex] \frac{f}{2} dT = -\frac{T}{V}dv [/tex]

    separation of variables because this is a differential equation

    [tex]
    \frac{f}{2} \frac{dT}{T} = -\frac{dV}{V} [/tex]

    integrating both sides with respect to initial and final volume/temperatures
    I get the equation

    [tex] \frac{f}{2} \int_{Ti}^{Tf} \frac{1}{T}dT = -\int_{Vi}^{Vf} \frac{1}{V}dV [/tex]

    integrating

    [tex] \left \frac{f}{2} lnT \right|_{Ti}^{Tf}= \left -lnV\right|_{Vi}^{Vf} [/tex]

    evaluating

    [tex] \frac{f}{2} lnTf - lnTi = -lnVf + lnVi [/tex]

    noting that lnx - lny is equivalent to ln(x/y)

    simplifying to get

    [tex] \frac{f}{2} ln\frac{Tf}{Ti} = ln\frac{Vi}{Vf} [/tex]

    this is where I was having the exponential problem

    exponentiating both sides

    do i raise e to the whole of the left side? or the individual parts?
    [tex] e^\frac{f}{2} e^{ln\frac{Tf}{Ti}} \textrm{ or do i } e^{\frac{f}{2} ln{Tf}{Ti} [/tex]
    what happens to the f/2?
    I know each of Tf and Ti should be raised to the power of f/2 afterwards

    but I do not understand how that works
     
  2. jcsd
  3. May 12, 2010 #2
    Quick info about exponentials:

    [tex]x^ax^b = x^{(a+b)}[/tex]

    [tex]\frac{x^a}{x^b} = x^{(a-b)} [/tex]

    [tex](x^a)^b = x^{ab} [/tex]

    So, when you exponentiate

    [tex] \frac{f}{2} ln\frac{Tf}{Ti} = ln\frac{Vi}{Vf} [/tex]

    You get

    [tex]e^{\frac{f}{2} ln\frac{T_f}{T_i}} = e^{ln\frac{V_i}{V_f}} [/tex]

    But remember that

    [tex]\frac{f}{2} ln\frac{T_f}{T_i} = ln(\frac{T_f}{T_i})^{(\frac{f}{2})}[/tex]

    Plug-in that on your equation to get

    [tex](\frac{T_f}{T_i})^{(\frac{f}{2})} = \frac{V_i}{V_f} [/tex]

    Was that what you wanted?

    ________________________

    EDIT:

    Instead, if you had done

    [tex]e^{\frac{f}{2}}e^{ln\frac{T_f}{T_i}} = e^{ln\frac{V_i}{V_f}} [/tex]

    On the left-hand side, you would be implying that

    [tex]e^{\frac{f}{2}}e^{ln\frac{T_f}{T_i}} = e^{(\frac{f}{2}+ln\frac{T_f}{T_i})} [/tex]

    Which is incorrect, as you can clearly see, because the original expression was

    [tex]\frac{f}{2}ln\frac{T_f}{T_i}[/tex]

    And not

    [tex]\frac{f}{2}+ln\frac{T_f}{T_i}[/tex]

    So a quick explanation to your question would be reviewing the first exponential property that I've posted above, that is, the product one.
     
    Last edited: May 12, 2010
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