Question

the estimated number of organisms in a population after t days is shown in the table below.

t days	estimated number of organisms, n
4	1250
6	1800
8	2600
10	3720

which equation best models the situation?
$t = 1.2(600)^n$
$n = 600(1.2)^t$
$n = 1.2(600)^t$
$t = 600(1.2)^n$

Explanation:

Step1: Recall exponential - growth formula

The general form of an exponential - growth model is $n = a(b)^t$, where $n$ is the quantity at time $t$, $a$ is the initial quantity, $b$ is the growth factor, and $t$ is the time.

Step2: Test the first data - point $(t = 2,n = 860)$ for each option

Option 1: $t=1.2(600)^n$ is not in the correct form of an exponential model for $n$ as a function of $t$.

Option 2: For $n = 600(1.2)^t$, when $t = 2$, $n=600\times(1.2)^2=600\times1.44 = 864$, which is close to 860.

Option 3: For $n = 1.2(600)^t$, when $t = 2$, $n=1.2\times(600)^2=1.2\times360000 = 432000$, which is not close to 860.

Option 4: $t = 600(1.2)^n$ is not in the correct form of an exponential model for $n$ as a function of $t$.

Step3: Test other data - points for the selected option

For $n = 600(1.2)^t$:
When $t = 4$, $n=600\times(1.2)^4=600\times2.0736 = 1244.16$, close to 1250.
When $t = 6$, $n=600\times(1.2)^6=600\times2.985984 = 1791.5904$, close to 1800.
When $t = 8$, $n=600\times(1.2)^8=600\times4.29981696 = 2579.890176$, close to 2600.
When $t = 10$, $n=600\times(1.2)^{10}=600\times6.1917364224 = 3715.04185344$, close to 3720.

Answer:

$n = 600(1.2)^t$