Question

over time, the number of organisms in a population increases exponentially. the table below shows the approximate number of organisms after y years.

y years	number of organisms, n
2	60
3	67
4	75

the 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?
12
24
61
82

Explanation:

Step1: Assume exponential - growth formula

The general form of an exponential - growth model is $n = ab^{y}$, where $n$ is the number of organisms, $y$ is the number of years, $a$ 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 $ab^{2}=60$, we get $\frac{55}{b}\times b^{2}=60$, which simplifies to $55b = 60$, so $b=\frac{60}{55}=\frac{12}{11}\approx1.0909$ and $a = 55\div\frac{12}{11}=55\times\frac{11}{12}=\frac{605}{12}\approx50.42$. So the exponential - growth model is $n=\frac{605}{12}\times(\frac{12}{11})^{y}$.

Step2: Solve for $y$ when $n = 600$

Set $n = 600$ in the equation $n=\frac{605}{12}\times(\frac{12}{11})^{y}$.
$600=\frac{605}{12}\times(\frac{12}{11})^{y}$.
First, rewrite it as $(\frac{12}{11})^{y}=\frac{600\times12}{605}=\frac{7200}{605}=\frac{1440}{121}$.
Take the natural logarithm of both sides: $y\ln(\frac{12}{11})=\ln(\frac{1440}{121})$.
Since $\ln(\frac{12}{11})\approx\ln(1.0909)\approx0.087$ and $\ln(\frac{1440}{121})\approx\ln(11.9008)\approx2.478$.
Then $y=\frac{\ln(\frac{1440}{121})}{\ln(\frac{12}{11})}=\frac{2.478}{0.087}\approx28.48$. The closest value among the options is 24.

Answer:

24