P(t)=10000e^.075t population at t years after 1980. Estimate year population doubled its 1980 population?

Question

Please show work. Thank you.

Answer 1

OK!

time (0) is 1980

So population in 1980 = p(0)= 10,000 e^(0.075*0) = 10,000

Therefore double the 1980 population is 20,000.

Now, solve for t:

20,000=10,000e^(0/075t)

2 = e^(0.075t)

ln 2 = 0.075t

(ln 2)/0.075 = t

0.693/0.075 = t

9.24 years = t

So the doubling year would have been in 1989, in late March.

Done.

Answer 2

2=e^.075t
.075t=ln 2
t=ln 2 /.075=9.24 years aprox 9 years 3 months

Answer 3

population in 1980 (t=0)

= 10,000*e^0.075t
= 10,000 * 1 (because e^0=1)
= 10,000

20,000 = 10,000 * e^0.075t (/10,000)
2 = e^0.075t
ln 2 = 0.075t
0.69314718056 = 0.075t
t = 9.24196240747