00:01
Okay, so we're giving our problem above.
00:03
So our first part is we're going to define k.
00:06
So define k, we're going to take the 61 .2 pounds of beef.
00:12
And we have our initial amount of 64 .6, and e, and then k is our rate.
00:20
And in that time frame, it was eight years.
00:24
Right.
00:24
So now we're going to use this equation to solve for k.
00:27
So we're going to divide both sides by 64 .6.
00:37
To 8k.
00:38
Here we're going to introduce natural log.
00:42
By introducing natural log here, that allows us to bring our exponent to the numerator because that's a natural log property.
00:50
So now we get the natural log of 61 .2 over 64 .6.
00:59
Now it equals 8k times the natural log of e.
01:04
Natural log and e are inverses of each other.
01:06
So that should be equivalent to 1.
01:09
So now only thing we had to do is divide both sides by 8.
01:13
So we should get a k value of negative 0 .00675...