If the rate of growth #P# is proportional to itself, then with respect to time #t#,
#[dP]/dt=kP#, ....inverting both sides, .....#dt/[dP]=[1]/[kP# and so integrating both sides
#intdt=int[dP]/[kP#, thus,..... #t=1/klnP +# a constant............#[1]#
Suppose #P# is some value # C# when# t=0#, substituting
#0=1/klnC+# constant, therefore the constant #= -1/klnC# and so substituting this value for the constant in ...#[1]# we have ,
#t= 1/k[ln P-lnC]# = #1/k ln[P/C]#, therefore , #kt=ln[p/C]#[ theory of logs] and so
#e^[kt]=P/C#......giving # P=Ce^[kt#. The constant #k# will represent the excess of births over deaths or vice versa for a decreasing rate.