Question

A population of bacteria is 8500 on Day 1, 9350 on Day 2, and 10285 on Day 3.

Answer 1

Answer: 134836

Step-by-step explanation:

The exponential growth equation is given by :-

$y=Ab^x$                              (1)

, where A= initial population

b= multiplicative growth arte.

x= Time period.

Given : A population of bacteria is 8500 on Day 1, 9350 on Day 2, and 10285 on Day 3.

i.e. A= 8500

$b=\dfrac{9350}{8500}=1.1$  [Multiplicative growth rate is the common ration between two terms.]

Put value of A and b in (1) , we get

$y=8500(1.1)^x$                     (2)

For Day 30

When x= 30-1=29            [∵Day 1 is the initial day.]

Put x= 29 in (2), we get

$y=8500(1.1)^{29}=8500(15.8630929717)\\\\=134836.290259approx134836$        [Round to the nearest whole number.]

Hence, the population be on Day 30 will be 134836 .

Answer 2

Answer:

The population on 30th day will be 134836.

Step-by-step explanation:

This is a case of exponential growth. The general form of exponential function is,

$y=ab^x$

where, a and b are constants.

The data points from the question are $(1,8500),(2,9350),(3,10285)$

Putting the values in the function,

for $x=1, y=8500$

$8500=ab$  ----------1

for $x=2, y=9350$

$9350=ab^2$  ------2

Dividing equation 2 by 1,

$\Rightarrow \dfrac{ab^2}{ab}=\dfrac{9350}{8500}=1.1$

$\Rightarrow b=1.1$

Putting the value of b in equation 1,

$\Rightarrow a=\dfrac{8500}{1.1}=7727.27$

Now the function becomes,

$y=7727.27(1.1)^x$

Putting x=30, we get

$y=7727.27(1.1)^{30}=134836.2\approx 134836$