Question

When cancer cells are subjected to radiation treatment, the proportion of cells that survive the treatment is
given by....

All Info in Next Post:

Answer 1

When cancer cells are subjected to radiation treatment, the proportion of cells that survive the treatment is
given by
\[P=e ^{-kr}\]
where r is the radiation level (measured in Rontgen) and k is a constant. If 40% of the cancer cells survive
when r = 500 Rontgen, what should the radiation level be in order to allow only 1% of cancer cells to survive?

Answer 2

so far:\[P=e ^{-0.6*500}\] is what I am thinking to work out the 40% left over. But stuck on how to proceed or if I am correct on this path.

Answer 3

Not quite. You are given P and r, use this to find k:\[0.4=e^{-500k}\]Taking the log of both sides lets us get k:\[\ln 0.4=-500k\]so,\[k \approx 0.002\]

Answer 4

Now that you have k, you can solve the next question. In the second part, you have k=0.002 and P=0.01 and need r=?

Answer 5

I'll set it up and let you solve it:\[0.01=e^{-0.002r}\]

Answer 6

I am unsure at this point what to do, would I repeat the same steps used to find k earlier but to now find r?

Answer 7

I am still lost.

Answer 8

take the natural log of both sides

Answer 9

ln(0.01) = -4.605170...
ln(-0.002) = -6.214608....

I've done that.

Answer 10

\[\ln 0.01=-0.002r\]\[-4.605...=-0.002r\]so,\[r=\frac{-4.605}{-0.002}\approx2300\]So to kill 99% of the cancer cells would take a radiation intensity of about 2300 Roentgen.

Answer 11

If the answer in your book is slightly different it is because k=0.002 is a rounded quantity. It is best to keep the unrounded values through the entire calculation and then round the final answer.

Answer 12

Thank you eseidl, I'm currently learning logs and how to use them so this application is new to me but certainly something I will continue to use. :) Cheers.