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Sagot :
[tex]Approximately\:6\:years[/tex]
1) Since the initial population of rabbits consists of 200 animals and there is a constant growth of 30% so we can start out writing the exponential model for this problem, like this.
[tex]P=200(1+0.3)^t[/tex]2) Let's plug P=1000 for the final population and solve for t:
[tex]1000=200(1+0.3)^t[/tex]So, let's count 2015 as our first year since we don't know when there will be 1,000 rabbits.
[tex]\begin{gathered} 1000=200(1.3)^t \\ \frac{1000}{200}=\frac{200(1.3)^t}{200} \\ 5=(1.3)^t \\ \ln(5)=\ln(1.3)^t \\ \ln \left(5\right)=t\ln \left(1.3\right) \\ t=\frac{\ln \left(5\right)}{\ln \left(1.3\right)} \\ t=6.13\approx6 \end{gathered}[/tex]Note the property of the exponents of logarithms
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