finding variables.

**thatsnotalion** · May 26th 2010, 09:54 PM

This could possibly be the second last math question i ever need help with =)

P=e^-kr

when...
P= proportion of cells which survive
r= radiation level
k= a constant

if 40% of cells survive (P), then r=500. when P=1% what would 'r' equal?

any help please?

**pickslides** · May 26th 2010, 10:11 PM

$0.4=e^{-500k}$

Now solve for $k$

**thatsnotalion** · May 26th 2010, 10:16 PM

hmmm k= ln0.4/500
therefore k= -1.83*10^-3?

**pickslides** · May 26th 2010, 10:30 PM

$0.4=e^{-500k}$

$-500k=\ln(0.4)$

$k=\frac{-\ln(0.4)}{500}$

**thatsnotalion** · May 26th 2010, 10:39 PM

hmmm ok. so when p=0.01 and k= (-ln0.4)/500r r=2512.9?

is that what you got?

**pickslides** · May 26th 2010, 10:45 PM

$P=e^{-kr}$

$k=\frac{-\ln(0.4)}{500}$

$P=e^{-\frac{-\ln(0.4)}{500}r}$

$P=e^{\frac{\ln(0.4)}{500}r}$

$P=0.01$

$0.01=e^{\frac{\ln(0.4)}{500}r}$

$\ln(.01)=\frac{\ln(0.4)}{500}r$

$\frac{\ln(.01)}{\frac{\ln(0.4)}{500}}=r$

$r=\frac{\ln(.01)}{\frac{\ln(0.4)}{500}}$

I get what ever that equals.

**thatsnotalion** · May 26th 2010, 10:53 PM

thanks. yeah that's the same answer. you're so smart!

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