# Homework Help: Logarithms disinfectant spray problem

1. Dec 8, 2007

### XJellieBX

Question:
A new disinfectant spray is expected to kill 50% of the known germs in a room, but for health reasons it can only be used once a day. Between spraying, the germs increase by 25%. How many consecutive days of spraying are required to reduce the germs in the room to 10% of the original amount?

Relevant equations:
A=Ao(1+i)^n

Attempt:
0.10Po=Po(0.50)^d + 0.25Po(0.50)^d
0.10=0.50^d + 0.25(0.50)^d
0.10=1.25(0.50)^d
0.08=0.50^d
log0.08=log0.50^d
log0.08=dlog0.50
d=$$\frac{log0.08}{log0.50}$$
d=4

Can someone please tell me what I'm doing wrong? The answer is supposed to be 5 days.

2. Dec 8, 2007

### rocomath

what do the variables of your relevant equation pertain to?

3. Dec 8, 2007

### XJellieBX

A is the final amount. Ao is the initial amount. i is the increase or decrease. and n is the number of periods.

4. Dec 8, 2007

### rocomath

i'm unable to follow your problem but when we solve for n, which i assume is the number of days

$$A=A_{0}(1+i)^{n}$$

divide by A initial and then take the log of both sides and then solve for n

5. Dec 8, 2007

### XJellieBX

yea, i tried doing that. but the answer didn't come out right.

6. Dec 8, 2007

### Dick

If A is the number of germs then after spraying A->0.5*A. After waiting a day that number increases by 25%. So that's multiplication by 1.25. Put the two together and from one day to the next A->0.5*A*1.25=0.625*A. So that's 0.1=(0.625)^n.