Question

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

A)12
B)24
C)61
D)82

Answer 1

Answer:

The correct answer is 29; the closest from the multiple choices given is B)24.

Step-by-step explanation:

Exponential functions are of the form

$y=a\times b^x$,

where a represents the initial population, b represents the percentage of growth as a decimal plus 1 (to represent keeping 100% of the original population) and x is the number of time periods.

We do not have the percentage of growth or the original population.  We can, however, write two equations in two variables to represent the information we  are given.

For the first year, we know the total to be 55.  We would then have the equation

$55=a\times b^1$

For the second year, we know the total to be 60.  We would then have the equation

$60=a\times b^2$

We want to isolate one of the variables in order to use substitution to solve this system.  Using the first equation, we will solve for a.  We do this by dividing both sides by b:

$\frac{55}{b}=\frac{a\times b}{b}\\\\\frac{55}{b}=a$

We can now substitute this for a in the second equation:

$60=a\times b^2\\\\60=(\frac{55}{b})\times b^2\\\\60=\frac{55b^2}{b}\\\\60=55b$

Divide both sides by 55:

$\frac{60}{55}=\frac{55b}{55}\\\\\frac{60}{55}=b$

We can now use this in the first equation to solve for a:

$55=a\times b^1\\\\55=a\times b\\\\55=a\times(\frac{60}{55})$

Divide both sides by 60/55 (remember when you divide by a fraction, you multiply by the reciprocal):

$55\times \frac{55}{60}=a\\\\\frac{55\times 55}{60}=a\\\\\frac{3025}{60}=a\\\\50.4167=a$

This gives us the exponential equation

$y=50.4167\times(\frac{60}{55})^x$

We are solving for the number of years (time periods) that it would take for the population to reach the maximum, 600.  Substitute 600 in for y:

$600=50.4167\times(\frac{60}{55})^x$

Divide both sides by 50.4167:

$\frac{600}{50.4167}=\frac{50.4167\times (\frac{60}{55})^x}{50.4167}\\\\11.9008=\frac{60}{55}^x$

We will use logarithms to solve this:

$\log_{\frac{60}{55}}11.9008=x\\\\28.46=x$

This means at 28 years, the population will not be at the max; at 29 years, it will be over.

Answer 2

We write the generic form equation:
 y = A (b) ^ t
 We must use two points in the table to find the constants A and b
 We have then:
 For: t = 1, y = 55
 55 = A (b)
 For: t = 2, y = 60
 60 = A (b) ^ 2
 We solve the system of equations:
 We divide both equations:
 (60 = A (b) ^ 2) / (55 = A (b))
 ((60) / (55)) = b
 b = 1.091
 We substitute in any of the equations:
 55 = A (b)
 55 = A (1,091)
 A = (55) / (1,091)
 A = 50.41 
 The equation that models the problem is:
 y = 50.41 * (1.09) ^ t
 For t = 29 we have:
 y = 613.5996987 (> 600)
 Answer:
 Assuming the trend continues the environment will not last until the population after 24 years.
 Answer:
 B) 24