precalc question

Question

The count in a bacteria culture was 389 after 10 minutes and 9517 after 40 minutes. Assuming the count grows exponentially, find the following:

The initial size of the culture:    

The culture's doubling time:    minutes (round to the nearest tenth)

The time when the count will reach 12000:    minutes (round to the nearest whole minute)

Answer 1

Let's do this problem entirely from scratch, deriving the equation for the count as a function of time. I will also do the math showing a lot more work.

Let C be the count, and let T be the elapsed time, in minutes.

The count C is a function of time, so we can write C(T).

The count is an exponential function of time, so the most general equation is C(T) = AB^T ,where A and B are constants that we need to figure out. It would be good to express A and B in terms of C(0) and D, where C(0) is the original count, and D is the doubling time.

At T = 0, C(T) = C(0) = AB⁰ = A, so A = C(0), and C(T) = C(0)B^T

At T = D, the count has doubled, so C(D) = 2C(0) = C(0)B^D. Dividing through by C(0), we get the equation 2 = B^D. We can take the 1/D power of both sides of this equation, so 2^1/D = B^D/D = B, giving us an expression for B.

Therefore, C(T) = C(0)2^T/D is the equation for the count as a function of T expressed in minutes.

In this problem, C(10) = C(0)2^10/D = 389, and C(40) = C(0)2^40/D = 9517, so we have 2 equations with 2 unknowns:

1. C(0)2^10/D = 389
2. C(0)2^40/D = 9517

We can solve for C(0) and D.

If we divide the second equation by the first equation, we get the equation C(0)2^40/D / [C(0)2^10/D] = 9517 / 389.

In the last equation, C(0) appears in the numerator and denominator of the left-hand side, and they cancel out, giving us 2^40/D / 2^10/D = 9517 / 389 = 2^40/D 2^-10/D = 2^{40/D -10/D} = 2^(40-10)/D = 2^30/D = 9517 / 389.

Taking the logarithm (with any base, it does not matter which) of both sides, log(2^30/D) = log(9517 / 389) = 30log(2)/D.

Solving for D, D = 30log(2)/log(9517 / 389) = 30log(2)/[log(9517)-log(389)] = 6.5038. Rounding to one decimal place, D = 6.5 is the doubling time in minutes.

To get C(0), we can go back to the equation C(0)2^10/D = 389, and solve for C(0) in terms of D:

C(0) = 389(2^-10/D) = 389(2^-10/6.5038) = 133.997 = 134.

So C(T) = C(0)2^T/D = 134*2^T/6.5.

C(T) = 12000 = 134*2^T/6.5 can be solved for T. Taking the logarithm (with any base, it does not matter which) of both sides, log(12000) = log(134) + (T/6.5)*log(2).

Solving for T, T = [log(12000) - log(134)]*6.5/log(2) = 42.15 = 42 minutes.

Answer 2

A is function of t, t in min.

A(10)=389=A₀r¹⁰

A(11)=A₀r¹¹

...

A(40)=9517=A₀r⁴⁰

r=(9517/389)^(1/30)

A₀=398/r¹⁰

A(t)=A₀r^t ,,,,,t≥0

for doubling, want A(t2)/A(t1)=2

ie,,,2=r^(t2-t1)

t2-t1=doubling interval=(ln2)/(ln(r)) which is not equal to ln(2/r)

12000=A₀r^t

solve for t