Question 1

${ } ^ { 14 } \mathrm { C }$ has a half life of 5730 years. How old is a piece of charcoal which has lost 90% of its ${ } ^ { 14 } \mathrm { C }$ ?
A) 20,000 years
B) 11,109 years
C) 9100 years
D) 19,109 years
E) none of these

Accepted Answer

D

Question 2

A radioactive substance is observed to disintegrate at a rate such that $\frac { 9 } { 10 }$ of the original amount remains after one year. What is the half-life of the substance?
A) 0.301 yr
B) 0.588 yr
C) 0.556 yr
D) 6.579 yr

Accepted Answer

The decay constant $\lambda$ can be found using the formula $N(t) = N_0 e^{-\lambda t}$. Given that $\frac{9}{10} = e^{-\lambda \cdot 1}$, solving for $\lambda$ gives $\lambda = -\ln(0.9)$. The half-life $T_{1/2}$ is given by $T_{1/2} = \frac{\ln(2)}{\lambda}$. Substituting $\lambda$ gives $T_{1/2} \approx 6.579$ years.

Question 3

A country has a population of 287 million in 2005. Assuming a growth rate of 1.3%, determine the function that expresses the population of the country t years after 2005.
A) P(t) = 287 $e ^ { - .013t }$
B) P(t) = 287 $e ^ { .013t }$
C) P(t) = 287 $e ^ { 1.3 t }$
D) P(t) = 287 $e ^ { - 1.3t }$
E) none of these

Accepted Answer

The correct function to model population growth with a constant rate is $P(t) = P_0 e^{rt}$, where $P_0$ is the initial population, $r$ is the growth rate, and $t$ is time in years. Therefore, with a growth rate of 1.3% or 0.013 as a decimal, the function is $P(t) = 287e^{0.013t}$.

Question 4

Plutonium has a decay rate of 0.003% per year. What is the half life?
A) 2310 years
B) 23,100 years
C) 33,000 years
D) 1630 years

Accepted Answer

The half-life of a substance is the time it takes for half of the initial amount to decay. To find the half-life of plutonium, we can use the formula:
t1/2 = (ln 2) / λ
where λ is the decay constant, which is equal to the decay rate expressed as a fraction (0.003/100 = 0.00003), and ln 2 is the natural logarithm of 2 (0.693). Substituting gives:
t1/2 = 0.693 / 0.00003 = 23,100 years
Therefore, the answer is B.

Question 5

Assume that a culture of bacteria grows at a rate proportional to its size such that if $10 ^ { 6 }$ bacteria are present initially, then there are 2 × $10 ^ { 6 }$ bacteria present after 3 hours. Determine a formula for the number of bacteria present after t hours in terms of powers of 2. (Hint: Recall that $b ^ { x }$ = $\mathrm { e } ^ { ( \ln b ) x }$ .)Enter your answer exactly in the form $\mathrm { P } ( \mathrm { x } ) = \mathrm { a }^ b \cdot 2 ^{\mathrm { c } / \mathrm { d }}$

Accepted Answer

The answer of Assume that a culture of bacteria grows...

Question 6

In a certain country, the rate of increase of the population is proportional to the population P(t). In fact, $P ^ { \prime } ( t ) = 0.23 P ( t )$ Suppose that initially the country's population is 50,000, and that 10 years later there are 500,000 people. Which of the following equations expresses this information mathematically?
A) 10 = $e ^ { 2.3 t }$
B) 500 = $e ^ { 2.3 ( 50 ) }$
C) 500,000 = $\mathrm { e } ^ { 0.23 ( 10 ) }$
D) 500,000 = 50,000 $e ^ { 2.3 }$
E) none of these

Accepted Answer

The answer of In a certain country, the rate of...

Question 7

The size of an insect colony t days after its formation is P( t) = 1000 $e ^ { 0.2 t }$ . Approximately how many insects are present after 10 days?
A) 690
B) 7389
C) 54,598
D) 6900
E) none of these

Accepted Answer

To find the size of the insect colony after 10 days, substitute $t = 10$ into the given equation: $P(10) = 1000e^{0.2 \times 10} = 1000e^2$. Evaluating this, $e^2 \approx 7.389$, so $1000 \times 7.389 = 7389$.

Question 8

A bacterial culture grows exponentially; that is, P(t) = 100 $\mathrm { e^{kt} }$ , where P(t) is the size of the culture at time t hours. Suppose that after 2 hours the size of the culture is 400. What is k (approximately)?
A) 3
B) 0.69
C) 0.06
D) 0.48
E) none of these

Accepted Answer

To find $k$, use the given information $P(2) = 400$ and the formula $P(t) = 100e^{kt}$. Substituting the values gives $400 = 100e^{2k}$. Dividing both sides by 100 gives $4 = e^{2k}$. Taking the natural logarithm of both sides gives $ln(4) = 2k$, so $k = \frac{ln(4)}{2} \approx 0.69$.

Question 9

Suppose that a school of fish in a pond grows according to the exponential law $\mathrm { P } ( \mathrm { t } ) = \mathrm { P } _ { 0 } \mathrm { e } ^ { \mathrm { kt } }$ and suppose that the size of the colony triples in 24 days. Determine k.
Enter your answer exactly in the form $\frac { \ln a } { \mathrm {~b} }$ where a, b are integers.

Accepted Answer

The answer of Suppose that a school of fish in...

Question 10

Radioactive carbon 11 has a half-life of 20 minutes. If there are 200 grams present at the start of our experiment, how many grams will remain after 10 minutes?
A) 141.421 g
B) 6.931 g
C) 50 g
D) 100 g

Accepted Answer

After 10 minutes, which is half of the half-life period, the amount of carbon 11 will be reduced to the square root of the initial amount due to the exponential decay process. The square root of 200 is approximately 141.421 grams.

Question 11

A parchment is offered for sale at a Paris flea market. The owner claims it is at least 2000 years old. However, a carbon-dating test shows that ${ } ^ { 14 } \mathrm { C }$ - ${}^{12} \mathrm { C }$ ratio for the manuscript is 95% of the corresponding ratio for currently manufactured parchment. How old is the manuscript? (The decay constant of ${ } ^ { 14 } \mathrm { C }$ is 0.00012)Enter your answer exactly in the form $\frac { \ln a } { \mathrm {~b} }$ (no units).

Accepted Answer

The answer of A parchment is offered for sale at...

Question 12

A certain radioactive substance is decaying at a rate proportional to the amount present. If 100 grams decays to 13.5 grams in 4 years, how long will it take for 90 grams to decay to 30 grams?
A) 2.405 yr
B) 2.195 yr
C) 0.501 yr
D) Problem cannot be solved as stated.

Accepted Answer

Let the amount present at time t be given by A(t). Then, we have:

A(0) = 100 (initial amount)
A(4) = 13.5 (amount after 4 years)

Since the decay is proportional to the amount present, we have:

A'(t) = -kA(t)

where k is the proportionality constant. To solve for k, we use the fact that A(4) = 13.5:

A'(4) = -kA(4)
(13.5 - 100)/4 = -k(13.5)
k = (100 - 13.5)/4(13.5) = 0.0139

Now, we want to find the time it takes for A(t) to decay from 90 to 30 grams:

A(t) = 90
A(0) = 100

We can solve for t by integrating:

ln|A(t)/A(0)| = -kt

ln|90/100| = -0.0139t

t = ln(10/9)/0.0139 ≈ 2.195 years

Therefore, the answer is B, 2.195 yr.

Question 13

A colony of bacteria is growing at a rate proportional to the number of bacteria present. At the beginning of an experiment there were about $10 ^ { 3 }$ bacteria present. In two hours, the count rose to $3 \times 10 ^ { 3 }$ bacteria. At what time will there be 6 × $10 ^ { 3 }$ bacteria present?
Enter just a real number rounded up to one decimal place (no units).

Accepted Answer

The answer of A colony of bacteria is growing at...

Question 14

Suppose that at any time t, a colony of fruit flies is growing at a rate equal to one half the current size of the colony. Find a formula which gives the size of the colony at time t if there were originally 500 fruit flies present.
Enter your answer exactly in the form: $\mathrm { P } ( \mathrm { t } ) = a \mathrm { e } ^ { \mathrm { b } t }$

Accepted Answer

The answer of Suppose that at any time t, a...

Question 15

Suppose that a school of fish in a pond grows according to the exponential law $P ( t ) = P 0 e ^ { k t }$ and suppose that the size of the colony triples in 24 days. If the initial size of the school was 50, when will the school contain 200 fish? Enter your answer exactly in the form $\frac { \mathrm { a } \ln \mathrm { b } } { \ln \mathrm { c } }$

Accepted Answer

The answer of Suppose that a school of fish in...

Question 16

A certain radioactive element has a half-life of 12 minutes. At what time is the substance decaying at a rate of 3.466 grams per minute if there are 120 grams present initially?
A) t = 0
B) t = 0.693 min
C) t = 0.058 min
D) t = 12 min

Accepted Answer

The decay rate is maximum at the beginning and decreases over time. At t = 12 min, the decay rate is calculated using the formula: rate = (ln(2)/half-life) * initial amount. Substituting the values gives the rate as approximately 3.466 grams per minute.

Question 17

Let P(t) be the quantity of strontium-90 remaining after t years. Suppose the half-life of strontium-90 is 28 years. Which of the following equations expresses the half-life information?
A) P(28) = $\frac { 1 } { 2 }$ $\mathrm { P } _ { 0 }$
B) P $\frac { 1 } { 2 }$ = 28 $\mathrm { P } _ { 0 }$
C) 28 = $\frac { 1 } { 2 }$ $\mathrm { P } _ { 0 }$
D) 28 = $\frac { 1 } { 2 }$ $\mathrm { P } _ { 0 }$ $\mathrm { e } ^ { - \mathrm { kt } }$
E) none of these

Accepted Answer

The half-life of a substance is the time it takes for half of it to decay. For strontium-90 with a half-life of 28 years, after 28 years, half of the original amount ($P_0$) would remain. Thus, $P(28) = \frac{1}{2}P_0$ correctly represents this situation.

Question 18

The population of a certain region was 10 million in 1950. By 1970, it had increased to 13.5 million. Assuming exponential growth, estimate the population in the year 2000.
Enter your answer exactly in the form a $\mathrm { e } ^ { \mathrm { C } }$ (no units).

Accepted Answer

The answer of The population of a certain region was...

Question 19

The population of a colony of bacteria triples in 3 days. Assuming that the rate of growth is proportional to the size of the population, how long did it take for the colony to double in size?
A) 2 days
B) 1.9 days
C) 4.9 days
D) 6 days
E) none of these

Accepted Answer

Let P be the population of the colony and t be the time it takes for the population to double. 
Then we can write: 
2P = P * e^(r*t), where r is the growth rate. 
Since the population triples in 3 days, we know: 
3P = P * e^(r*3), which simplifies to: 
e^(r*3) = 3. 
Taking the natural logarithm of both sides gives: 
r * 3 = ln(3), so r = ln(3)/3. 
Substituting this value of r into the first equation yields: 
2P = P * e^[(ln(3)/3)*t], which simplifies to: 
2 = e^[(ln(3)/3)*t]. 
Taking the natural logarithm of both sides gives: 
ln(2) = (ln(3)/3)*t, so t = (3/ln(3)) * ln(2) ≈ 1.9 days.

Question 20

Sixteen pounds of a radioactive substance loses one fourth of its original mass in 2 days. The mass m(t) remaining at time t is given by:
A) m(t) = 16 $e ^ { - 0.693t }$
B) m(t) = 16 $e ^ { 0.693t }$
C) m(t) = 4 $e ^ { - 0.693t }$
D) m(t) = 16 $e ^ { 0.144t }$

Accepted Answer

The answer of Sixteen pounds of a radioactive substance loses...

Quiz 5: Applications of the Exponential and Natural Logarithm Functions

${ } ^ { 14 } \mathrm { C }$ has a half life of 5730 years. How old is a piece of charcoal which has lost 90% of its ${ } ^ { 14 } \mathrm { C }$ ?

A radioactive substance is observed to disintegrate at a rate such that $\frac { 9 } { 10 }$ of the original amount remains after one year. What is the half-life of the substance?

A country has a population of 287 million in 2005. Assuming a growth rate of 1.3%, determine the function that expresses the population of the country t years after 2005.

Plutonium has a decay rate of 0.003% per year. What is the half life?

The size of an insect colony t days after its formation is P( t) = 1000 $e ^ { 0.2 t }$ . Approximately how many insects are present after 10 days?

A bacterial culture grows exponentially; that is, P(t) = 100 $\mathrm { e^{kt} }$ , where P(t) is the size of the culture at time t hours. Suppose that after 2 hours the size of the culture is 400. What is k (approximately) ?

Radioactive carbon 11 has a half-life of 20 minutes. If there are 200 grams present at the start of our experiment, how many grams will remain after 10 minutes?

A certain radioactive substance is decaying at a rate proportional to the amount present. If 100 grams decays to 13.5 grams in 4 years, how long will it take for 90 grams to decay to 30 grams?

A certain radioactive element has a half-life of 12 minutes. At what time is the substance decaying at a rate of 3.466 grams per minute if there are 120 grams present initially?

Let P(t) be the quantity of strontium-90 remaining after t years. Suppose the half-life of strontium-90 is 28 years. Which of the following equations expresses the half-life information?

The population of a certain region was 10 million in 1950. By 1970, it had increased to 13.5 million. Assuming exponential growth, estimate the population in the year 2000. Enter your answer exactly in the form a $\mathrm { e } ^ { \mathrm { C } }$ (no units).

The population of a colony of bacteria triples in 3 days. Assuming that the rate of growth is proportional to the size of the population, how long did it take for the colony to double in size?

Sixteen pounds of a radioactive substance loses one fourth of its original mass in 2 days. The mass m(t) remaining at time t is given by: