# Interesting exponential/logarithm example

• Mar 12th 2008, 11:41 PM
bobchiba
Interesting exponential/logarithm example
One of the more budding students in my class derived this example question, I think his dad works in a hostpital, and its based on actual paperwork on MRSA in hospitals...anyway, If anyone can give me a hand as Im useless with exponentials..

- Hostpital rooms can be decontaminated of MRSA if sprayed with hydrogen peroxide, however this is harmful to people if there's prolonged exposure, so we must know its decay rate.
- A room was sprayed with vapour containing hydrogen peroxide, with a concentration of 22.80 mmoles/litre, and the level of this peroxide, ct, was measured over intervals:

t (min)----------0--------5-------10-------20------
Ct (mmoles/ltr) 22.80 -- 17.60 -- 13.80 -- 8.25

As far as I can conclude, the decay equation is ct = coe^-λt

a) how do you tranform the equation so the straight line of y=mx+c can be plotted and how are y, m and c related to ct, co, and λ
• Mar 12th 2008, 11:52 PM
Moo
Hello,

$C_t=C_0 e^{- \lambda t}$

Generally, for getting the linear form, you transform all this stuff in logarithms.
The graphic will be ln(Ct)=f(t), so the ordinate will be ln(Ct) and the abscissa t.

To transform it, compose with the logarithm :

$\ln(C_t) = \ln(C_0 e^{- \lambda t})$

But we know that $\ln(ab) = \ln(a) + \ln(b)$ and that $\ln(e^x) = x$

So $\ln(C_t)$ will be $\ln(C_0) - \lambda t$
• Mar 13th 2008, 05:07 AM
bobchiba
Thanks MOO, although I admit I probably sound like I know more than I do. I grasp the concept of what you've layed down, but I don't really have much experience of going through the motions, If anyone could develop on whats there or have any other examples, would be great,
• Mar 13th 2008, 07:53 AM
Moo
Well,

What you can remember is that the logarithm transforms a product into a sum, because of the ln(ab)=ln(a)+ln(b) propriety =)

You want something in the y=mx+c form. Here, t is x. y will be ln(Ct), not Ct as you can think on first sight. It's common, you have to remember that usually, you will draw a line with ln(Ct) in ordinates and t in abscissa.

$\ln(C_t) = \ln(C_0) + \ln(e^{- \lambda t}) = \ln(C_0) - \lambda t$
This is a line i didn't put previously, but it explains all, am sorry, but i don't really see what i can explain more :/

y=mx+c

Here, m = $- \lambda$ and c is $\ln(C_0)$ ;)
• Mar 13th 2008, 08:20 PM
bobchiba
how would I find the value of the decay constant...ie, λ
• Mar 13th 2008, 11:44 PM
Moo
With the values given in the text ;)

You've been given t and $C_t$, so you just need two equations to find $C_0$ and $\lambda$