00:01
The initial population and we're trying to find the double time.
00:04
So we are told that at 10 hours the bacteria was at 5 ,000 and then at 12 hours we're told that the bacteria was at 6 ,000 and that it's proportional.
00:20
So we're going to set up a proportion.
00:22
We're going to do 5 ,000 over 6 ,000 is equal to social population.
00:30
Times e to the 10k power, so time was 10 for that top part, and then the initial population times e to the 12k for 12 hours at the 6 ,000.
00:48
So the 5 ,000 over 6 ,000 is going to reduce to 5 over 6.
00:52
The initial populations are going to cancel out for now, and we can divide e to the 10k over the 12k by subtracting the exponent.
01:01
So 10 minus 12 is negative 2.
01:04
So we're going to have e to the negative 2k.
01:07
So we're going to raise both sides to the natural log in order to get the e and the natural log to cancel.
01:15
We're going to divide both sides by negative 2.
01:18
And what this accomplishes is that gives us our k, which we need in order to find the initial population and the double time.
01:27
So natural log of 5, 6, maybe you close that parenthesis in your calculator...