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Math Help - logarithm of concentration

  1. #1
    Newbie DavePercy's Avatar
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    logarithm of concentration

    When dealing with reaction rates for first order reactions, my textbook has this formula:

    \ln [A_f] = -kt + \ln [A_0]

    where [A_f],[A_0] are the final and initial concentrations or pressures of a reactant, and k is a constant with units of 1/sec. This is derived from

    \frac{[A_f]}{[A_0]} = e^{-kt}

    Is there any meaning in asking what the logarithm of a unit of concentration or pressure is? In this case it looks like the inputs to ln are actually dimensionless since the units cancel in the second equation above. In general though, what does a logarithm or exponential function output when given a unit, like \ln(20 mi/hr), e^{12 Joules}, or \ln(5 ArbitraryUnits)? Do we have to leave it as something like a real number plus ln(mi) - ln(hr) ?
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  2. #2
    o_O
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    Yes, generally when dealing with logarithms, they deal with unitless operations (exception is when looking at pH where pH = -log[H^{+}]. The connotations in doing these operations are implied).

    Looking at first-order reactions, you know that the equilibrium constant k has a unit of s^{-1}. So, rearranging as you did:

    \ln\left(\frac{[A]_{t}}{[A]_{0}}\right) = -kt

    Inside the operation of the logarithm, the units cancel out and on the right hand side, the units cancel out as well (given that time is in seconds).
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  3. #3
    Newbie DavePercy's Avatar
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    Is there any situation where real-world units are passed to logarithms or exponents?
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