Thread: finding variables.

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?

$\displaystyle 0.4=e^{-500k}$

Now solve for $\displaystyle k$

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

$\displaystyle 0.4=e^{-500k}$

$\displaystyle -500k=\ln(0.4)$

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

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

is that what you got?

$\displaystyle P=e^{-kr}$

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

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

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

$\displaystyle P=0.01$

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

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

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

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

I get what ever that equals.

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