00:01
Hey, it's clarissa enumerate here.
00:03
So we get dp over d t is equal to c ln of m over p times p.
00:19
To get one over p, l n m over p, d p is equal to c d t.
00:38
We're going to integrate both sides and we get negative du is equal to 1 over pdp.
00:52
How we got this is that we make u be equal to ln of m over p.
01:03
So we got du, it's equal to the derivative of m over p over p, which we get to be to be negative 1 over p dp.
01:24
Now we're just going to substitute to get negative 1 over u.
01:31
We integrate this, du, is equal to the integral of cd t to get negative ln of the ln of m over p is equal to c t plus d.
01:52
We know that m is bigger than p and an ln of a number bigger than one is going to be positive so we just don't need the absolute volume signs now to get l n of ln of m of p over p is equal to negative c t minus d we put the base to the base of e to get e to the negative c t of my minus d we get ln of m minus ln of p, because this is, you can subtract it if it's a quotient, and we get negative ct minus d.
02:45
We end up getting ln of m minus ln of p is equal to a, e to the negative c t, ln of m minus a, e to the negative c t, ln of m minus a, e to the negative c t, is equal to ln of p...