Log Help (Maths, not Forestry)
Posted: Tue Nov 29, 2022 10:09 am
Hi All,
I find myself needing to use logs for the first time in years and something is going on which I don't understand. The problem I have is based on Beer's Law (transmission)
I(z) = Io * e^(-az) -> Eqn(1)
Where I(z) is irradiance at after some z through an absorbing thing, Io is the starting irradiance (set to one), a is absorption co-eff, and z is the distance travelled.
I want to find the absorption needed to drop the transmission to 50% after 10mm...
Therefore I(z)/Io = 0.5, and z = 10 and I need to find the value of a...
So...
I(z) / Io = 0.5 = e^(-10a)
So ln(0.5) = -10a and therefore a = -ln(0.5) / 10 = 0.0693... (note sign)
The problem is when I sub this back into Eqn(1) as a sanity check... (setting Io as 1)
I(z) = e^(-az) = e^(-0.0693*10) = e^(-0.693) = 2, which is not 0.5
I know that e^(0.693) = 0.5
Where does the sign change come from in the exponent? There seems to be an inconsistency here that I can't track down...
I find myself needing to use logs for the first time in years and something is going on which I don't understand. The problem I have is based on Beer's Law (transmission)
I(z) = Io * e^(-az) -> Eqn(1)
Where I(z) is irradiance at after some z through an absorbing thing, Io is the starting irradiance (set to one), a is absorption co-eff, and z is the distance travelled.
I want to find the absorption needed to drop the transmission to 50% after 10mm...
Therefore I(z)/Io = 0.5, and z = 10 and I need to find the value of a...
So...
I(z) / Io = 0.5 = e^(-10a)
So ln(0.5) = -10a and therefore a = -ln(0.5) / 10 = 0.0693... (note sign)
The problem is when I sub this back into Eqn(1) as a sanity check... (setting Io as 1)
I(z) = e^(-az) = e^(-0.0693*10) = e^(-0.693) = 2, which is not 0.5
I know that e^(0.693) = 0.5
Where does the sign change come from in the exponent? There seems to be an inconsistency here that I can't track down...