(2)
Integrating both sides of equation (2), we get
log P =
C1
20
t +
or
P =
C1
et20
e⋅
or
P =
20
C
et
 (where 
C1
C
e
=
) (3)
Now
P = 1000,   when t = 0
Substituting the values of P and t in (3), we get C = 1000 Therefore, equation (3),
gives
P = 1000 
t20
e
Let t years be the time required to double the principal Then
2000 = 1000
et20
    ⇒  t = 20 loge2
EXERCISE 9