SOLUTION: The population of Adamsville grew from 8000 to 15000 in 6 years. Assuming uninhibited exponential​ growth, what is the expected population in an additional 2 ​years?

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Question 1139101: The population of Adamsville grew from 8000 to 15000 in 6 years. Assuming uninhibited exponential​ growth, what is the expected population in an additional 2 ​years?
Found 2 solutions by josmiceli, greenestamps:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+P+=+P%5B0%5D%2A%28+1+%2B+r+%29%5Et+
+15000+=+8000%2A%28+1+%2B+r+%29%5E6+
+1.875+=+%28+1+%2B+r+%29%5E6+
Take the log of both sides
+.273+=+6%2Alog%28+1+%2B+r+%29+
+.0455+=+log%28+1+%2B+r+%29+
+10%5E.0455+=+1+%2B+r+
+1.11045+=+1+%2B+r+
+r+=+.11045+
--------------------------
+P+=+15000%2A%28+1+%2B+.11045+%29%5E2+
+P+=+15000%2A1.2331+
+P+=+18496.59+ in 2 more years
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check answer:
+P+=+8000%2A%28+1+%2B+.11045+%29%5E8+
+P+=+8000%2A2.31202+
+P+=+18496.18+
( error due to rounding off )

Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


The other tutor showed a very formal mathematical solution to your problem.

Here is a less formal solution which gets you to the answer much faster and with far less work.

The growth factor over 6 years was 15000/8000 = 15/8.

Assuming a constant rate of exponential growth, the growth factor over a period of 2 years will be (15/8)^(2/6) or (15/8)^(1/3).

So the population after another 2 years will be

15000%2815%2F8%29%5E%281%2F3%29+=+18496.59 to 2 decimal places.

So the population will be about 18497.