Question

 Question:

A certain type of bacteria doubles in population every hour. Their growth can be modeled by an exponential equation. Initially, there are 15 bacteria colonies present in the sample. 

A. Find an equation to model the bacteria growth.
B. How many colonies are present after 3 and a half hours?
C. When will there be 584 colonies present?

Answer 1

A. If the population is doubling every hour the base of the exponential is 2 to the power of x. Then, the initial population is 15, so when X is o Y has to be 15. 15 is the beginning number. we get this equation:
Y=15(2^x)

B. Plug 3.5 into X:
Y=15(2^3.5)=(about) 170

C. Since it is asking what X value will 584 bacteria be present, it now gives us Y and wants us to solve for X:
584=15(2^x)
*divide by 15*
584/15=2^x
*log on both sides to get the X out of the exponent position*
log(584/15)=log(2^x)
Move X in front of log
log(584/15)=Xlog(2)
*Divide by log(2)*
Log(584/15)/log(2)=X=(about) 5.28 hours

Answer 2

A. Equation to model the bacteria growth is:

$y=15(2^x)$

B. Approximately 170 colonies are present after 3 and a half hours.

C. After 5.28 hours there will be 584 colonies present.

Given that a certain type of bacteria doubles in population every hour.

Initially, there are 15 bacteria colonies present in the sample.

Now,

A. If the population is doubling every hour the base of the exponential is 2 to the power.

Since the initial population is 15

That is

when x=0 ,y = 15. 15

we get the equation as:

$y=15(2^x)$

Now,

B. Put x=3.5 into the equation

$y(3.5)=15(2^3.5)\\=170$

Now,

C. y= 584 colonies

We find x here:

So,

584=15(2^x)

Taking log on both sides we get  

 log(38.93)=log(2^x)

log(38.93)=x log(2)

log(38.93) ÷log(2)=x

x = 5.28 hours

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