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...

## Log Help (Maths, not Forestry)

### Log Help (Maths, not Forestry)

You can't polish a turd...

unless its Lion or Osterich poo... http://dsc.discovery.com/videos/mythbus ... -turd.html

unless its Lion or Osterich poo... http://dsc.discovery.com/videos/mythbus ... -turd.html

- shpalman
- Princess POW
**Posts:**7567**Joined:**Mon Nov 11, 2019 12:53 pm**Location:**One step beyond-
**Contact:**

### Re: Log Help (Maths, not Forestry)

I definitely get e^(-0.693) = 0.5 and e^(0.693) = 2, and it has to be that way around because e is 2.71828... while any number^0 is 1, so for e^x to be greater than 1 but less than 2.71828, x needs to be positive but less than 1, and for e^x to be less than 1, x needs to be less than 0 i.e. negative.

*having that swing is a necessary but not sufficient condition for it meaning a thing*

@shpalman@mastodon.me.uk

- shpalman
- Princess POW
**Posts:**7567**Joined:**Mon Nov 11, 2019 12:53 pm**Location:**One step beyond-
**Contact:**

### Re: Log Help (Maths, not Forestry)

(double post)

*having that swing is a necessary but not sufficient condition for it meaning a thing*

@shpalman@mastodon.me.uk

### Re: Log Help (Maths, not Forestry)

Yep, e^(-0.693) = 0.5 and e^(0.693) = 2.

Jaap's Page: https://www.jaapsch.net/

### Re: Log Help (Maths, not Forestry)

Yes, I think it's whatever you're using to calculate e^(-0.693) is not applying the negation to the exponent properly

My avatar was a scientific result that was later found to be 'mistaken' - I rarely claim to be 100% correct

ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn

Meta? I'd say so!

ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn

Meta? I'd say so!

### Re: Log Help (Maths, not Forestry)

Ah jeez - embarrassing calculator usage fail...

I'd stored the answer in memory without accounting for the sign change and then was using that result (with the wrong sign) in the sanity check...

Thanks all!

I'd stored the answer in memory without accounting for the sign change and then was using that result (with the wrong sign) in the sanity check...

Thanks all!

You can't polish a turd...

unless its Lion or Osterich poo... http://dsc.discovery.com/videos/mythbus ... -turd.html

unless its Lion or Osterich poo... http://dsc.discovery.com/videos/mythbus ... -turd.html