Given function : y = 5.5 (1.025)×, Where exponent x is number of years from now and y give the population of rabbits in hundreds.
We need to find the time(s) when there will be 600 rabbits and 1200 rabbits.
Solution: We know, y represents the population of rabbits in hundreds.
a) Plugging y=600 in given function, we get
600 = 5.5 (1.025)×
Dividing both sides by 5.5
600/5.5 = 5.5 (1.025)× / 5.5
600/5.5 = (1.025)×
Taking natrual log on both sides, we get
ln (600/5.5) = ln (1.025)×
ln (600/5.5) = x ln (1.025)
Dividing both sides by ln(1.05), we get
![\frac{ln\frac{600}{5.5}}{ln(1.05)}=x](https://tex.z-dn.net/?f=%5Cfrac%7Bln%5Cfrac%7B600%7D%7B5.5%7D%7D%7Bln%281.05%29%7D%3Dx)
x= 96.171 years apprimately.
b) Plugging y=1200 in given function, we get
1200 = 5.5 (1.025)×
Dividing both sides by 5.5
1200/5.5 = 5.5 (1.025)× / 5.5
1200/5.5 = (1.025)×
Taking natrual log on both sides, we get
ln (1200/5.5) = ln (1.025)×
ln (1200/5.5) = x ln (1.025)
Dividing both sides by ln(1.05), we get
![\frac{ln\frac{1200}{5.5}}{ln(1.05)}=x](https://tex.z-dn.net/?f=%5Cfrac%7Bln%5Cfrac%7B1200%7D%7B5.5%7D%7D%7Bln%281.05%29%7D%3Dx)
x= 110.377 years apprximately.
Therefore, it will take 96.171 years apprimately to population to reach 600 rabbits and 110.377 years approximately to population to reach 1200 rabbits.