SOLUTION: On a small island there were 32768 deer and only 128 wolves. Statistics gathered revealed that the wolf population doubled every 3 years and the deer population halved every 6 year

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Question 1181923: On a small island there were 32768 deer and only 128 wolves. Statistics gathered revealed that the wolf population doubled every 3 years and the deer population halved every 6 years. How many years did it take for the wolf population to overtake the deer population? Create an algebraic solution to the problem.
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
wolves=w
deer=d
t=years
w=wo*2^(t/3)
=128*2^(t/3)
--
d=do(1/2)^(t/6)
=32768*(2)^(-t/6)
w=2^7*2^(t/3)=2^(7+(t/3))
d=2^15*2^(-t/6)=2^(15-(t/6))
-
7+(t/3)=15-(t/6)
t/2=8
t=16 years
---
128*2^(16/3)=5160.63
32768*0.5^(16/6)=5160.63

Answer by ikleyn(52795) About Me  (Show Source):
You can put this solution on YOUR website!
.
On a small island there were 32768 deer and only 128 wolves. Statistics gathered revealed that
the wolf population doubled every 3 years and the deer population halved every 6 years.
How many years did it take for the wolf population to overtake the deer population?
Create an algebraic solution to the problem.
~~~~~~~~~~~~~~~ 

From the condition, we have 

    D(t) = 32768*1/2)^(t/6)     (1)


for deer population and

    W(t) = 128*2^(t/3)          (2)


for wolves population.


To answer the question, we should solve this equation

    D(t) = W(t),   or  32768*(1/2)^(t/6) = 128*2^(t/3).    (3)


It implies

    32768%2F128 = 2^(t/3) * 2^(t/6),

or

    256 = 2^(t/3 + t/6)

    2^8 = 2^(2t/6 + t/6)

    2^8 = 2^(3t/6) = 2^(t/2).


Hence,  t%2F2 = 8  and  t = 2*8 = 16  years.      ANSWER

Solved.