P(t) = 500 / [1+20,000e^(-0.549t) ]
How fast means rate of change of the function, and that is its derivative.
Then, find the derivative of P(t), P '(t), using the chain rule:
P '(t) = - 500 / [ 1 + 20,000e^(-0.549t)]^2 ] * 20,000*(-0.549)e^(-0.549)t
P '(t) = 5,490,000 * e^ (-0.549t) / [1 + 20,000e^-0,549t)]^2
P'(t) = 5,490,000 * e^(0.549t) / [ e^(0.549t) + 20,000]^2
Now replace t by 10 .
P'(10) =5,490,000 * e^(5.49) / [ e^(5.49) + 20,000]^2 = 3.25 years
Answer: 3.3 years