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.
162
likes809 views
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
4.5105 Answers
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.
Frequently asked questions (FAQs)
What is the median of a data set with 13 elements?
+
What is the surface area of a cube with side length s?
+
Question: What is the derivative of a function f(x) if f(x) is the integral from a to x of g(t) dt?
+
New questions in Mathematics