SOLUTION: The population of a certain species of a fish has a relative growth rate of 1.2% per year. It is estimated that the population in the year 2000 was 10 million. If the fish pop

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: The population of a certain species of a fish has a relative growth rate of 1.2% per year. It is estimated that the population in the year 2000 was 10 million. If the fish pop      Log On


   

Question 1151357: The population of a certain species of a fish has a relative growth rate of 1.2% per year. It is estimated that the population in the year 2000 was 10 million.

If the fish population has reached about 13 million how many years have passed?
The equation is N(t)=Ne^rt
Found 2 solutions by josmiceli, ikleyn:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In the year 2000, you can say +t+=+0+
and +N%280%29+=+10+ in millions
+N%28t%29+=+N%2Ae%5E%28r%2At%29+
+N%280%29+=+N%2Ae%5E%28.012%2A0%29+
+10+=+N%2Ae%5E0+
+N+=+10+ at +t=0+
--------------------------------
+13+=+10%2Ae%5E%28+.012%2At+%29+
+e%5E%28+.012t+%29+=+1.3+
+.012t+=+ln%28+1.3+%29+
+.012t+=+.2624+
+t+=+21.86+
About 22 yrs have passed when population
reaches 13 M

Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.

The "hint" about the form of the equation in the post is "perpendicular" (or "opposite") to the logic of the problem.


The form of the equation, consistent with the post, is THIS exponential function


    N(t) = N*1.012^t,


where "t" is the time in years after 2000 and N = 10 millions (in the year of 2000).


Then you need to find the time "t" in years from the equation


    13 = 10*1.012^t.


Divide both sides by 10


    13%2F10 = 1.012^t,

    1.0.12^t = 1.3


and take the logarithm base 10 from both sides


    t*log(1.012) = log(1.3),

    t = log%28%281.3%29%29%2Flog%28%281.012%29%29 = 21.995 years = 22 years (approximately).     ANSWER

Solved, answered and explained.