# Thread: Interesting exponential/logarithm example

1. ## 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

2. Hello,

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

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 :

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

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

So $\displaystyle \ln(C_t)$ will be $\displaystyle \ln(C_0) - \lambda t$

3. 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,

4. 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.

$\displaystyle \ln(C_t) = \ln(C_0) + \ln(e^{- \lambda t}) = \ln(C_0) - \lambda t$
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 = $\displaystyle - \lambda$ and c is $\displaystyle \ln(C_0)$

5. how would I find the value of the decay constant...ie, λ

6. With the values given in the text

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