dave13
carcass
Four hours from now, the population of a colony of bacteria will reach 1.28 * \(10 ^ 6\). If the population of the colony doubles every 4 hours, what was the population 12 hours ago?
A. 6.4 * \(10 ^ 2\)
B. 8.0 * \(10 ^ 4\)
C. 1.6 * \(10 ^ 5\)
D. 3.2 * \(10 ^ 5\)
E. 8.0 * \(10 ^ 6\)
pushpitkc, is my approach / reasoning correct ?
1.28 * \(10 ^ 6\) means we need to move decimal point 6 points to the right so it is 1,280,000, so i just leave out zeros and work with 128
if in 4 hours the population will be 128, then now it is 64
4 hours ago it was 32
8 hours ago 16
12 hrs ago 8
so all i need is to add 4 zeros and express it like this \(8*10^4\)

one thing I didn't get why in answer option B. 8.0 * \(10 ^ 4\) \(8\) is expressed as \(8.0\) and not just \(8\) I mean with one zero after decimal point...

Hi
dave13This problem is best solved when we go backwards!
8 *10^4 = 80000
Now see the sequence when we double the population of the colony of bacteria
80000 * 2 = 160000 | 160000 * 2 = 320000 | 320000 * 2 = 640000 | 640000 * 2 = 1280000
(You can see that 12 hours from the start, the population has increased
from 8 *10^4 to 6.4 *10^5 and then to 1.28 * 10^6)
Your approach is perfect otherwise. All you need to understand is this part of why the ZEROES increase!
Hope this helps you!