1. Find the doubling time and amount in 30 years if the initial deposit was $1000 and the annual rate is 8.6%.
2. Find the initial deposit and doubling time if the annual rate is 5.25% and the amount in 30 years is 10,405.37.
3. Suppose that the cholera bacteria in a colony grows unchecked according to the Law of Exponential Change. The colony starts with 1 bacterium and doubles in number every half hour. How many bacteria will the colony contain at the end of 24 hours?
4. A sample of Ce-143 (an isotope of cerium) loses 99% of its radioactive matter in 199 hours. What is the half life of Ce-143?
2. Find the initial deposit and doubling time if the annual rate is 5.25% and the amount in 30 years is 10,405.37.
3. Suppose that the cholera bacteria in a colony grows unchecked according to the Law of Exponential Change. The colony starts with 1 bacterium and doubles in number every half hour. How many bacteria will the colony contain at the end of 24 hours?
4. A sample of Ce-143 (an isotope of cerium) loses 99% of its radioactive matter in 199 hours. What is the half life of Ce-143?
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1)
2P = P * (1 + 0.086)^t
2 = 1.086^t
ln(2) = t * ln(1.086)
t = ln(2) / ln(1.086)
t = 8.4016596101098791178689971430495
A = 1000 * (1.086)^30
A = 1000 * 11.882142530420574616566507285708
A = 11882.142530420574616566507285708
2)
2 = 1.0525^t
ln(2) = t * ln(1.0525)
t = ln(2) / ln(1.0525)
t = 13.54642156234993831401236291443
10405.37 = P * 1.0525^30
10405.37 / 1.0525^30 = P
P = 2241.7872379251938905617588311723
3)
A = 1 * 2^(24 / (1/2))
A = 2^48
A = 281474976710656
4)
A = P * (1/2)^(t / k)
A = 0.01 * P
t = 199
Solve for k
0.01 * P = P * (1/2)^(199 / k)
1/100 = (1/2)^(199 / k)
-ln(100) = (199/k) * -ln(2)
k = 199 * ln(2) / ln(100)
k = 29.952484568566128923767020025087
2P = P * (1 + 0.086)^t
2 = 1.086^t
ln(2) = t * ln(1.086)
t = ln(2) / ln(1.086)
t = 8.4016596101098791178689971430495
A = 1000 * (1.086)^30
A = 1000 * 11.882142530420574616566507285708
A = 11882.142530420574616566507285708
2)
2 = 1.0525^t
ln(2) = t * ln(1.0525)
t = ln(2) / ln(1.0525)
t = 13.54642156234993831401236291443
10405.37 = P * 1.0525^30
10405.37 / 1.0525^30 = P
P = 2241.7872379251938905617588311723
3)
A = 1 * 2^(24 / (1/2))
A = 2^48
A = 281474976710656
4)
A = P * (1/2)^(t / k)
A = 0.01 * P
t = 199
Solve for k
0.01 * P = P * (1/2)^(199 / k)
1/100 = (1/2)^(199 / k)
-ln(100) = (199/k) * -ln(2)
k = 199 * ln(2) / ln(100)
k = 29.952484568566128923767020025087