Question

On January 1, 1990, 50 frogs are released at a lake that previously had no frogs. 12 years later, it is estimated that there are 12,000 frogs around the lake. Assume that the annual percentage increase in the frog population has been the same over this time period. Determine in which year there are one million frogs at the lake if the population continues to increase exponentially at the same rate.

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Answer to a math question On January 1, 1990, 50 frogs are released at a lake that previously had no frogs. 12 years later, it is estimated that there are 12,000 frogs around the lake. Assume that the annual percentage increase in the frog population has been the same over this time period. Determine in which year there are one million frogs at the lake if the population continues to increase exponentially at the same rate.

Neal
4.5
Låt tillväxtekvationen modelleras som y\left$t\right$=50a^t där t är antalet år efter 1990 givet att y$0$ = 50 och y$12$ = 12000 y$12$ = 12000 ger ekvationen som 50a^{12}=12000 \Rightarrow a^{12}=\frac{12000}{50} \Rightarrow a^{12}=240 \Rightarrow a=240^{\frac{1}{12}} Sätt y$t$ = 1 miljon = 1000000 för att få ekvationen som 50\left$240\right$^{\frac{t}{12}}=1000000 \Rightarrow\left$240\right$^{\frac{t}{12}}=\frac{1000000}{50} \Rightarrow\left$240\right$^{\frac{t}{12}}=20000 Ta logaritmer med bas 10 på båda sidor för att få \frac{t}{12}\left$\log240\right$=\log20000 \Rightarrow t=\frac{12\times\log20000}{\log240} \Rightarrow t\approx\frac{12\times4.30103}{2.38021}\approx21.68 Obligatoriskt år = 1990 + 21,68 = 2011,68 = 2012 $avrundat till närmaste kalenderår$ Svar: Ungefär år 2012 kommer det att finnas 1 miljon grodor.

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