over time, the number of organisms in a population increases exponentially. the table below shows the…
over time, the number of organisms in a population increases exponentially. the table below shows the approximate number of organisms after y years.\n| y years | number of organisms, n |\n| ---- | ---- |\n| 1 | 55 |\n| 2 | 60 |\n| 3 | 67 |\n| 4 | 75 |\nthe environment in which the organism lives can support at most 600 organisms. assuming the trend continues, after how many years will the environment no longer be able to support the population?\n12\n24\n61\n82
Answer
Answer:
12
Explanation:
Step1: Assume exponential - growth formula
Let the exponential - growth formula be $n = ab^{y}$, where $n$ is the number of organisms, $y$ is the number of years, and $a$ and $b$ are constants. Using the first two data points $(y = 1,n = 55)$ and $(y = 2,n = 60)$: When $y = 1$, $n=ab^{1}=55$, so $a\times b = 55$, then $a=\frac{55}{b}$. When $y = 2$, $n = ab^{2}=60$. Substitute $a=\frac{55}{b}$ into $ab^{2}=60$: $\frac{55}{b}\times b^{2}=60$, which simplifies to $55b = 60$, so $b=\frac{60}{55}=\frac{12}{11}$, and $a = 55\div\frac{12}{11}=55\times\frac{11}{12}=\frac{605}{12}$. The formula is $n=\frac{605}{12}\times(\frac{12}{11})^{y}$.
Step2: Solve for $y$ when $n = 600$
Set $n = 600$ in the formula $600=\frac{605}{12}\times(\frac{12}{11})^{y}$. First, rewrite it as $(\frac{12}{11})^{y}=\frac{600\times12}{605}=\frac{1440}{121}$. Take the natural - logarithm of both sides: $y\ln(\frac{12}{11})=\ln(\frac{1440}{121})$. Since $\ln(\frac{12}{11})\approx0.087$ and $\ln(\frac{1440}{121})\approx2.44$, then $y=\frac{\ln(\frac{1440}{121})}{\ln(\frac{12}{11})}=\frac{2.44}{0.087}\approx12$.