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)

Results 1 to 4 of 4

- Apr 13th 2010, 11:40 AM #1

- Joined
- Apr 2008
- Posts
- 123

- Apr 13th 2010, 09:13 PM #2

- Joined
- Mar 2010
- Posts
- 1,032
- Thanks
- 269

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

- Apr 14th 2010, 11:18 AM #3

- Joined
- Apr 2008
- Posts
- 123

- Apr 15th 2010, 12:51 AM #4

- Joined
- Mar 2010
- Posts
- 1,032
- Thanks
- 269

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:

, where k is the constant of proportionality. Note the negative sign - as x increases, r decreases.

The solution is . 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

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

or approximately 33.5mg.

It's essentially the same calculation as before.

- Hollywood