Apopulation of 550 rabbits is increasing by 2.5% each year the function y = 5.5 (1.025)× give the population of rabbits in hundreds x year from now about how long will it take the population to reach 600 rabbits? 1200 rabbits?

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

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

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.

entering that problem on the tube might get you an answer

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.