Exponential Functions math word problem

Question

A bacteria culture starts with 620620 bacteria and grows at a rate proportional to its size. After 44 hours there will be 24802480 bacteria.

 

1) Express the population P after t hours as a function of t .
P(t)=____________________

 

2)What will be the population after 3 hours? 

    ______________________

 

3)Estimate how long will it take for the population to reach 1890? Give your answer accurate to the nearest whole number.

    _________________________

Answer 1

Hi Courtney,

 

The formula you need is  Amount NOW = (Amount ORIGINALLY)(e^kt)

 

So we know the original amount, it is 620.  

 

The formula becomes:   Amount NOW = 620 e^kt

 

Now we use the fact that when t = 4, we have 2480 bacteria.   So we find "k" by plugging these numbers into our new formula.

 

2480 = 620 e^k(4)

 

Now we need to solve for k.   

Divide both sides by 620

4 = e^k(4)

Take the logarithm (ln) of both sides to "pull the k down" out of the exponent position.

 

ln(4) = 4k

 

so k = ln(4)/4  which is approximately   k =0.34657

 

Now you can use the finalized formula to answer all of the questions.

 

FINALIZED FORMULA:

 

Amount NOW = 620 e^{0.34657 t}   or, in terms of population P(t)

 

1.)  P(t) = 620 e^{0.34657 t}

 

2.)  P(3) = 620 e^(0.34657)(3) = 1753 bacteria

 

3.)  1890 = 620 e^{0.34657 (t)}   3.048387 = e^{0.34657 t}

 

ln(3.048387) = 0.34657 t

 

3.2161 = t    to the nearest hour,  would either be 3 hours or 4, not sure if your teacher will want you to round up or down.  I would think you would need the answer to be 4 hours, because after just 3 hours it has not yet reached 1890 bacteria.  But 4 hours it will be over 1890 bacteria.  

 

Answer 2

P(t) = P₀e^kt

 

P₀ = 620

t = 4

P(4) = 2480

 

2480 = 620e^k(4)

 

k = ln(2480/620) / 4

 

(1)

 

The function is

 

P(t) = 620e^kt,

 

where k = ln(2480/620) / 4

 

(2)

 

Plug t = 3 into P(t) and evaluate.

 

(3)

 

Set 1890 = P(t) and solve for t.