I with spectrophotometry ( The Beer-Lambert Law )

[BBoy.Goon]

I'm doing a science project about the antioxidant activity of red wines from my country. This days, I was going through the theoretical part ( the basics of spectrophotometry, photometric analysis ect. ) and I'm currently at the Beer-Lambert Law. I'm would like to know the whole mathematical path that is used so that starting from this two equations: A = 2 - log₁₀ %T and A = \epsilonbc,

where:
A is absorbance, ( no units )
T is transmittance, ( no units )
\epsilon is molar absorbtivity, ( L mol^-1cm^-1 )
b is the length of the sample ( path length of the cuvette ) ( cm ) and
c is the concentration of the compound in solution ( mol L^-1 )

the equation %T = e ^-\epsilonbc is deduced.

note: the molar absorbtivity is represented by the letter epsilon from the Greek alphabet and it's not superscripted as it looks like.

Thanks in advance

 

[Char. Limit]

Well, let's see...

Start by putting the two equations equal to each other...

2-log_{10}(T)=\epsilon b c

Add log_10(t) to both sides...

2=\epsilon b c + log_{10}(T)

Subtract epsilon b c from both sides...

2- \epsilon b c = log_{10}(T)

And raise both sides to the power of 10.

T=10^2 * 10^{-\epsilon b c}

I got a different answer than that... I'm not sure if that's right.

 

[BBoy.Goon]

I started just like you ( putting the two equations equal to each other ) but I think that e, from the equation %T = e^-\epsilonbc stands for the so called Neper's number ( e = 2,72 ) and after using the rule that:

log_ab = log_cb / log_ca

and several trasnformations I got stuck here:

log_e%T = 2log_e10 - \epsilonbc log_e10