# Thread: 80% to 95% - How much lysol will you need?

1. ## 80% to 95% - How much lysol will you need?

Spraying 18.0 mg of lysol will kill 80% of bacteria on the table. How much lysol must be sprayed to kill 95% of these bacteria on the table. (it will have an exponential relationsip)

2. If you spray 18mg, there are 20% of the bacteria remaining. If you spray another 18mg, it will kill all but 20% of the remaining bacteria, so there will be 20% * 20% = 4% remaining. Let r be the portion of bacteria remaining and x the amount of lysol sprayed. Then

r = (20%)^(x/18mg)

So we want to find the x that makes r=5%. As a check, since 36mg kills just a little too much, your answer should be slightly less than 36mg.

Post again in this thread if you're still having trouble.

- Hollywood

3. Originally Posted by hollywood
If you spray 18mg, there are 20% of the bacteria remaining. If you spray another 18mg, it will kill all but 20% of the remaining bacteria, so there will be 20% * 20% = 4% remaining. Let r be the portion of bacteria remaining and x the amount of lysol sprayed. Then

r = (20%)^(x/18mg)

So we want to find the x that makes r=5%. As a check, since 36mg kills just a little too much, your answer should be slightly less than 36mg.

Post again in this thread if you're still having trouble.

- Hollywood
Thanks. This is more clear.

How would we get the answer using differentials (like dP/dt)? and the exponential e?

4. Ok. Let r be the fraction of bacteria remaining, and let x be the amount of lysol sprayed. If you spray a very small amount, it will kill an amount of bacteria proportional to the fraction of bacteria remaining:

$\frac{dr}{dx}=-kr$, where k is the constant of proportionality. Note the negative sign - as x increases, r decreases.

The solution is $r=Ce^{-kx}$. Since not spraying any lysol results in no bacteria killed, r=1 when x=0, so C=1.

We also know that r=0.2 when x=18mg, so

$0.2=e^{-k(18mg)}$

$k=-\frac{\ln{0.2}}{18mg}$

So we want to know what x gives r=0.05.

$0.05=e^{-kx}$

$x=-\frac{\ln{0.05}}{k}=\frac{\ln{0.05}}{\ln{0.2}}(18m g)$ or approximately 33.5mg.

It's essentially the same calculation as before.

- Hollywood