Weber-Fecher law - logarithm problem

Hello everybody!

I've a problem reagrding the Weber-Fecher law (the law describes the relation between a objective stimulus and it's subjective perception).

It looks like that: E(R) = c * lg(R/R0)

E: subjective perception (i.e.: loudness in phon)
R: objective stimulus (i.e.: sonic intensity in W/m^2)
c : 1/k; k: Weber Constant (for hearing k ~ 0.09 => c=10)

My problem is that you can find the law written down both with lg and ln as logarithm.

The law is derived as follows:

dE = c * dR/R

after integration of both sides:

E = c * ln(R) + C

for integrating the dR/R my formula collection says that § (f'(x)/f(x) )dx = ln|f(x)| + C => § (dR/R) dR = ln|R| + C (§ should be the integral sign)

so in my opinion the version with the ln is right, but my school book says that it shoud be lg ! I think my error must be in the integration of dR/R ... there must be another way with a lg as result... :confused:

Suggestions welcome!

Thanks in advance,
Flo

P.S.: I hope you understand my problem. I'm german and it's difficult to find the right english counterparts for some mathematic expressions... so don't mind my bad english...;)

Raptor, in BASIC, QuickBASIC and VB, LOG stands for the natural logaritm. So, your book's "LG" must stand for natural logarithm, also, because your answer, with ln (meaning natural logartithm), is correct.

1) I think:
ln = log_e
lg = log₂
log = log₁₀

2) I'm not sure if this helps, but:
log_ab = log_cb / log_ca
(i think thats it)

In that case,
E(R) = c * lg(R/R₀)
becomes:
E(R) = c * ln(R/R₀) / ln(2) (<----or ln(10) if lg=base 10)

Thanks for your help!

It seems that both versions (either with log₁₀ = lg or log_e = ln) are possible.

I found this exerpt of a book also covering Fechner's law (on page 11 and 12):
http://media.wiley.com/product_data/...0471378887.pdf

They also get the solution with the natural logarithm.
But the author(s) also write that Fechner uses a mathematically not proven 'auxiliary principle' (as he called it) when deriving his law and that his law is only approximately true.

By the way IMO: log₂ a = lb a;


Regards,
Flo