exponential growth and decay help?

Question

Two bacteria colonies are cultivated in a laboratory. The first colony has a doubling time of 2 hours and the second a doubling time of 3 hours. initially, the first colony contains 1000 bacteria and the second colony 3,000 bacteria. At what time t will sizes of the colonies be equal??

I'd appreciate any help. Thanks.

Answer 1

well okay.
so you know the formula for decay and growth problems;
B(t) = Bo(b)^t/n

B = variable for bacteria
Bo = initial amount.
t = time
b = what is happening to it (in this case, it is doubling, therefore there would be a 2)
n = how often it happens.



so thennn you set up two different equations.
for the first one, it would be
B(t) = 1000(2)^t/2

and the second
B(t) = 3000(2)^t/3

now you can put those into your graphing calculator and see where the two lines intersect. use X as the variable for t.


Y1 = 1000(2)^t/2
Y2 = 3000(2)^t/3


I think you can do the rest and I hope that helped.

Answer 2

Let's set up the two equations. The first, P1 = 1000(e^rt) and the second, P2 = 3000(e^rt) where r is a constant that needs to be determined for both equations.

We know that 2000=(1000)e^2r for the first and that 6000=(3000)e^3r for the second. By taking the natural log, ln, of both sides of each of these equations, we can solve for r.

In the first, ln(2000) = ln(1000e^2r). Solving this yields r = .3466.

In the seond ln(6000) = ln(3000e^3r). Solving this yields r=.2315.

The time at which the two populations are equal will be when 1000e^.3466t = 3000e^.2315t.

Again, taking the natural log of both sides enables us to solve for t = 9.55 hours.

Answer 3

The equation is p(t)=p(0)e^rt
Colony 1, 2000=1000e^2r ,letting t=2, r = growth rate
2=e^2r
ln (2) = 2r
r=ln (2) /2
r=0.34657 for colony 1
Colony 2, 6000=3000e^3r ,letting t=3, r = growth rate
2=e^3r
ln (2) = 3r
r= ln (2) / 3 for colony 2
r=0.23105 for colony 2
equating colony 1 & 2
1000 e^0.34657t = 3000 e^0.23105t
e^0.34657t=3e^0.23105t
0.34657t = ln(3)+ 0.2310t
0.34657t=1.0986+0.2310t
t(0.34657-0.2310)=1.0986
0.11567t=1.0986
t=1.0986/0.11567
t=9.5 hours

Answer 4

sure, the exponent is the "t" being elevated by skill of the single million/5th. A(t) represents the quantity after time t. So 4 is the preliminary volume as a results of fact A(0) = 4. one million/2 skill to shrink by skill of one million/2 (with exponent t/5). i think of you're engaged on some questions approximately radioactive decay? If sure, then the single million/2-life is 5, as a results of fact whilst t = 5, A(t) = A(0) / 2, i.e., will become one million/2 of the preliminary volume. the two t and the "5" could have a unit, le us say in twelve months. So after 20 years, the finished volume left is A(20) = 4 * (one million/2) ^(20/5) = 4 * (one million/2) ^ 4 = one million/4.

Answer 5

2000 = 1000e^2k --> k = ln(2)/2,so
x = 1000 e^ln(2)/2t for first colony
x = 3000e^ln(2)/3t for 2nd colony
1000 e^ln(2)/2t = 3000e^ln(2)/3t
2000 e^t/2 = 6000 e^t/3
e^t/2 = 3e^t/3
e^t/6 = 3
t/6 = ln(3)
t = 6ln(3)