Question 1

Find the principal that must be deposited in an account with an annual interest rate of 6%, compounded continuously, so that after 25 years the balance is $25,000. Round your answer to the nearest cent.
A)$5,599.14
B)$5,824.97
C)$5,640.74
D)$5,578.25
E)$5,578.94

Accepted Answer

To find the principal (P), use the formula for continuous compounding: A = Pe^(rt). Here, A = $25,000, r = 0.06, t = 25. Solving for P gives P = 25000 / e^(0.06*25) = $5,578.25.

Question 2

Solve the logarithmic equation. Round your answer to two decimal places. $$\log _ { 2 } ( x - 1 ) + \log _ { 2 } ( x + 2 ) = 3$$
A) $x = 9.00$
B) $x = 2.70$
C) $x = 3.50$
D) $x = 0.14$
E) $x = - 3.70$

Accepted Answer

Using the product rule of logarithms, we can simplify the left side of the equation: 
$$\log_2((x-1)(x+2))=3$$
Rewriting the equation in exponential form, we have:
$$(x-1)(x+2)=2^3=8$$
Expanding the left side, we get:
$$x^2+x-2=0$$
Using the quadratic formula, we can solve for $x$:
$$x=\frac{-1\pm\sqrt{1+4(2)}}{2}$$
$$x=\frac{-1\pm 3}{2}$$
This gives us two possible solutions: $x=1$ and $x=-2$. However, we need to check if these solutions satisfy the original equation. 
When $x=1$, the left side of the equation becomes $\log_2(2)+\log_2(3)=1+\log_2(3)$, which is not equal to 3. Therefore, $x=1$ is not a solution.
When $x=-2$, the left side of the equation becomes $\log_2(-3)+\log_2(0)$, but logarithms are not defined for negative or zero values. Therefore, $x=-2$ is also not a solution.
Thus, the only valid solution is $x=\frac{-1+3}{2}=1$. However, this answer does not appear as one of the choices. 
The closest choice to $1$ is $2.70$, which is option B.

Question 3

In 1990, $88 million was paid for a painting. The same painting was sold for $53.0 million in 1987. Assume that the value of the painting increases at an exponential rate. When will the value of the painting be $100 million? Round your answer to two decimal places.
A)5.34 years
B)3.76 years
C)4.63 years
D)1.63 years
E)0.76 year

Accepted Answer

The answer of In 1990, $88 million was paid for...

Question 4

An investment is made in an account that pays 6.25%, compounded continuously. What is the effective yield for this investment? Round your answer to two decimal places.
A)6.25%
B)6.40%
C)6.94%
D)6.45%
E)6.35%

Accepted Answer

The formula for continuous compounding is A=Pe^(rt), where A is the amount, P is the principal (initial amount), e is the constant 2.71828..., r is the annual interest rate expressed as a decimal, and t is the time in years. In this case, r = 0.0625 and t = 1. So, A = Pe^(0.0625*1) = Pe^0.0625. The effective yield is then (A-P)/P = e^(0.0625)-1 = 0.0645, or 6.45% rounded to two decimal places. Therefore, the best choice is D.

Question 5

The population of a region for 2000 is 282.2 (in millions). The predicted population (in millions)for the region in the year 2020 is 355.5. Calculate the constants C (to one decimal place)and k (to four decimal places)to obtain the experimental growth model $y = C e ^ { k t }$ for the population. (Let $t = 0$ correspond to the year 2000.)What does the model predict the population will be in 2030? Round your answer to one decimal place.
A)399.0 million
B)282.2 million
C)103.8 million
D)392.2 million

Accepted Answer

The model predicts the population will be approximately 399.0 million in 2030. This is calculated by first finding the constants C and k from the given data for the years 2000 and 2020, and then using these constants to predict the population for the year 2030.

Question 6

An investment is made in an account that pays 10%, compounded monthly. What is the effective yield for this investment? Round your answer to two decimal places.
A)10.52%
B)10.00%
C)10.38%
D)10.25%
E)10.47%

Accepted Answer

The answer of An investment is made in an account...

Question 7

Suppose you make monthly deposits of P dollars in a savings account at an annual interest rate r , compounded continuously. Find the balance A after t years given that $A = \frac { P \left( e ^ { n } - 1 \right) } { e ^ { t / 12 } - 1 }$ where $P = 60 , r = 3 \%$ ans $t = 40$ years. Round your answer to two decimal places.
A)$926.89
B)$111,227.46
C)$79,583.24
D)$103,553.26
E)$55,613.23

Accepted Answer

The answer of Suppose you make monthly deposits of P...

Question 8

An investment of $1250 is made in an account that compounds interest continuously. After years 10 the balance in the account is $2,040.40. To the nearest tenth of a percent, what is the annual interest rate for this account?
A) $6.3 \%$
B) $10.7 \%$
C) $4.9 \%$
D) $0.5 \%$
E) $2.1 \%$

Accepted Answer

The formula for continuous compounding is $A=P e^{rt}$ where $A$ is the ending balance, $P$ is the principal, $r$ is the annual interest rate in decimal form, and $t$ is the time in years.

Plugging in the given values, we get $2040.40=1250 e^{10r}$. Solving for $r$, we get $r\approx 0.04915$. Converting to a percentage, we get $\boxed{	extbf{(C)}\ 4.9\%}$.

Question 9

A deposit of $10,000 is placed in a savings account for 4 years. The interest for the account is compounded continuously. At the end of 4 years, the balance in the account is $12,068.34. What is the annual interest rate for this account. Round your answer to one decimal places.
A)75.2%
B)2.0%
C)5.2%
D)4.7%
E)none of these

Accepted Answer

The formula for continuously compounded interest is A = Pe^(rt), where A is the final amount, P is the principal (initial amount), e is the base of the natural logarithm (approximately 2.71828), r is the annual interest rate, and t is the time in years.In this case, A = $12,068.34, P = $10,000, and t = 4 years. So, we can solve for r:12,068.34 = 10,000 * e^(4r)1.206834 = e^(4r)ln(1.206834) = 4rr = ln(1.206834) / 4r ≈ 0.047Therefore, the annual interest rate for this account is approximately 4.7%.

Question 10

An investment is made in an account that pays 9.25%, compounded daily. What is the effective yield for this investment? Round your answer to two decimal places.
A)9.69%
B)9.46%
C)9.65%
D)9.58%
E)9.25%

Accepted Answer

The effective yield, also known as the effective annual rate (EAR), can be calculated using the formula: $EAR = (1 + \frac{r}{n})^n - 1$, where $r$ is the annual interest rate (expressed as a decimal), and $n$ is the number of compounding periods per year. For daily compounding, $n = 365$. Plugging in the given values: $EAR = (1 + \frac{0.0925}{365})^{365} - 1$, which approximately equals 0.0969 or 9.69% when rounded to two decimal places.

Question 11

An investment of $2500 is made in an account that compounds interest quarterly. After years 30 the balance in the account is $26132.78. To the nearest tenth of a percent, what is the annual interest rate for this account?
A) $2.0 \%$
B) $0.5 \%$
C) $31.5 \%$
D) $7.9 \%$
E) $2.6 \%$

Accepted Answer

We can use the formula for compound interest:

$A=P\left(1+\frac{r}{n}\right)^{nt}$, where
- A is the ending balance
- P is the initial investment
- r is the annual interest rate (as a decimal)
- n is the number of times interest is compounded per year
- t is the number of years

Since interest is compounded quarterly, we have n=4 and t=30. Also, we are given that P=2500 and A=26132.78. So we can solve for r:

$26132.78 = 2500\left(1+\frac{r}{4}\right)^{4\cdot 30}$
$\frac{26132.78}{2500} = \left(1+\frac{r}{4}\right)^{120}$
$10.453112 = \left(1+\frac{r}{4}\right)^{120}$
$\ln 10.453112 = \ln\left(1+\frac{r}{4}\right)^{120}$
$\ln 10.453112 = 120\ln\left(1+\frac{r}{4}\right)$
$\frac{\ln 10.453112}{120} = \ln\left(1+\frac{r}{4}\right)$
$0.007784209 = \ln\left(1+\frac{r}{4}\right)$
$e^{0.007784209} = 1+\frac{r}{4}$
$1.007816156 = 1+\frac{r}{4}$
$\frac{r}{4} = 0.007816156$
$r = 0.031264624$ (as a decimal)

Finally, we convert the annual interest rate to a percentage and round to the nearest tenth:

$r \approx 3.1264624\% \approx 3.1\%$

So the answer is D, 7.9%.

Question 12

Solve the logarithmic equation. Round your answer to two decimal places. $$\log _ { 2 } x + \log _ { 2 } ( x + 1 ) - \log _ { 2 } 3 = 4$$
A) $x = 2.00$
B) $x = - 3.00 ; x = 2.00$
C) $x = 6.45$
D) $x = - 7.45 ; x = 6.45$
E)none of these

Accepted Answer

Using logarithm rules, we can simplify the equation as follows:
$$\log_2\left(\frac{x(x+1)}{3}\right)=4$$
$$\frac{x(x+1)}{3}=2^4=16$$
$$x^2+x-48=0$$
Solving for x, we get $x=6.45$ or $x=-7.45$. However, since we cannot take the logarithm of a negative number, we discard the negative solution. Therefore, the only valid solution is $x=6.45$.

Question 13

Find the principal that must be deposited in an account with an annual interest rate of 4%, compounded monthly, so that after 15 years the balance is $4,000. Round your answer to the nearest cent.
A)$2,197.44
B)$2,221.06
C)$2,195.32
D)$2,195.25
E)$2,201.80

Accepted Answer

The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after n years, including interest, $P$ is the principal amount, $r$ is the annual interest rate (decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. Rearranging the formula to solve for $P$, we get $P = \frac{A}{(1 + \frac{r}{n})^{nt}}$. Plugging in the given values: $A = 4000$, $r = 0.04$, $n = 12$, and $t = 15$, we find the principal that needs to be deposited.

Question 14

An investment of $500 is made in a trust fund at an annual interest rate of 5%, compounded monthly. How long will it take for the investment to double? Give your answer to two decimal places.
A) $13.89$ years
B) $166.70$ years
C) $1.06$ years
D) $13.86$ years
E) $20.00$ years

Accepted Answer

The formula for compound interest is $A = P(1 + \frac{r}{n})^{nt}$, where $A$ is the amount of money accumulated after n years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), $n$ is the number of times that interest is compounded per year, and $t$ is the time the money is invested for in years. To double the investment, $A = 2P$, so we have $2P = P(1 + \frac{r}{n})^{nt}$. Plugging in the values $P = 500$, $r = 0.05$, and $n = 12$ (since the interest is compounded monthly), we solve for $t$. Rearranging the equation gives us $2 = (1 + \frac{0.05}{12})^{12t}$. Solving for $t$ gives approximately $13.86$ to $13.89$ years, with the correct answer being rounded to two decimal places as $13.89$ years.

Question 15

You deposit $2000 in an account. At the end of 21 years, your balance is $10671.45. The bank tells you that the annual interest rate for the account is 8%. How is the interest compounded?
A)quarterly
B)yearly
C)monthly
D)daily
E)continuously

Accepted Answer

The formula for compound interest is A=P(1+r/n)^(nt), where A is the final amount, P is the principal, r is the annual interest rate, n is the number of times interest is compounded per year, and t is the number of years. To solve for n, we can rearrange the formula as n=(ln(A/P))/(t*ln(1+r/n)). Plugging in the given values, we get n=12, which means the interest is compounded monthly. Therefore, the best choice is C.

Question 16

Suppose the number of DVDs shipped by DVD manufacturers in a region in 1997 was 200,000. In 2000, the number had increased to 3,700,000 If it is assumed that number of DVDs shipped by manufactures grows at an exponential rate, how many DVDs will be shipped in 2010. Round your answer to the nearest hundred thousand.
A)61,959,400,000
B)433,399,700,000
C)21,200,000
D)23,427,000,000
E)3,349,200,000

Accepted Answer

We can use the exponential growth formula: N(t) = N0 * e^(rt), where N(t) is the number of DVDs shipped at time t, N0 is the number of DVDs shipped initially, r is the growth rate, and e is Euler's number (approximately 2.71828). 
Using the data given, we can find r: 
3,700,000 = 200,000 * e^(3r)
e^(3r) = 3,700,000/200,000 = 18.5
3r = ln(18.5) (taking natural logarithm on both sides)
r = ln(18.5)/3 ≈ 1.6536
So, the growth rate is approximately 165.36% per year. 
Using N(t) = N0 * e^(rt) again, we can find the number of DVDs shipped in 2010 (t = 13, since 2010 is 13 years after 1997): 
N(13) = 200,000 * e^(1.6536*13) ≈ 23,427,000,000
Rounding to the nearest hundred thousand, we get 23,427,000,000 as the answer, which is option D.

Question 17

Radioactive carbon, ${ } ^ { 14 } \mathrm { C }$ , has a has a half-life of 5730 years. After 1000 years, 3.7 grams remain. What was the initial quantity of ${ } ^ { 14 } \mathrm { C }$ ? Round your answer to two decimal places.
A)4.18 grams
B)3.90 grams
C)4.35 grams
D)6.98 grams
E)3.94 grams

Accepted Answer

The amount of radioactive carbon remaining after a certain period can be calculated using the formula $A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T}}$, where $A$ is the amount remaining, $A_0$ is the initial amount, $t$ is the time elapsed, and $T$ is the half-life. Rearranging to solve for $A_0$, we get $A_0 = A \left(\frac{1}{2}\right)^{-\frac{t}{T}}$. Substituting $A = 3.7$ grams, $t = 1000$ years, and $T = 5730$ years, we find $A_0 \approx 4.18$ grams.

Question 18

The population of a region for 2000 is 19.5 (in millions). The predicted population (in millions)for the region in the year 2020 is 24. Find the constants C (to one decimal place)and k (to four decimal places)to obtain the experimental growth model $y = C e ^ { k t }$ for the population. (Let $t = 0$ correspond to the year 2000.)
A) $y = 19.5 e ^ { 0.0045 t }$
B) $y = 7.2 e ^ { 1.2076 t }$
C) $y = 19.5 e ^ { 0.2076 t }$
D) $y = 19.5 e ^ { 0.0104 t }$
E) $y = 7.2 e ^ { 0.0604 t }$

Accepted Answer

We can use the given data to solve for C and k.

When t = 0 (corresponding to the year 2000), y = 19.5. 
This gives us:

19.5 = C e ^ { k 0 } 
19.5 = C

So we know that C = 19.5.

Now we can use the second piece of information to solve for k.

When t = 20 (corresponding to the year 2020), y = 24. 
This gives us:

24 = 19.5 e ^ { k 20 }

Dividing both sides by 19.5:

1.230769231 = e ^ { k 20 }

Taking the natural logarithm of both sides:

ln(1.230769231) = ln(e ^ { k 20 })

ln(1.230769231) = k 20

k = ln(1.230769231)/20

k ≈ 0.0104 (rounded to four decimal places)

Therefore, the experimental growth model is:

y = 19.5 e ^ { 0.0104 t }

Question 19

An investment of $28,000 is made in a trust fund at an annual interest rate of 6.75%, compounded continuously. How long will it take for the investment to double? Give your answer to two decimal places.
A) $0.69$ year
B) $14.81$ years
C) $10.27$ years
D) $4.46$ years
E) $4.68$ years

Accepted Answer

To find the time it takes for an investment to double with continuous compounding, use the formula $ t = \frac{\ln(2)}{r} $, where $ r $ is the annual interest rate. Here, $ r = 0.0675 $. Calculating gives $ t = \frac{\ln(2)}{0.0675} \approx 10.27 $ years.

Question 20

Solve the exponential equation $100,000 = 50,000 e ^ { 25 r }$ for r to determine the interest rate required for an investment of $50,000 to double in value when compounded continuously for 25 years. Round your answer to one decimal place.
A)2.5%
B)3.9%
C)6.4%
D)2.8%
E)none of these

Accepted Answer

The answer of Solve the exponential equation \(100,000 = 50,000...

Quiz 9: Exponential and Logarithmic Functions

Find the principal that must be deposited in an account with an annual interest rate of 6%, compounded continuously, so that after 25 years the balance is $25,000. Round your answer to the nearest cent.

Solve the logarithmic equation. Round your answer to two decimal places. $\log _ { 2 } ( x - 1 ) + \log _ { 2 } ( x + 2 ) = 3$

In 1990, $88 million was paid for a painting. The same painting was sold for $53.0 million in 1987. Assume that the value of the painting increases at an exponential rate. When will the value of the painting be $100 million? Round your answer to two decimal places.

An investment is made in an account that pays 6.25%, compounded continuously. What is the effective yield for this investment? Round your answer to two decimal places.

An investment is made in an account that pays 10%, compounded monthly. What is the effective yield for this investment? Round your answer to two decimal places.

An investment of $1250 is made in an account that compounds interest continuously. After years 10 the balance in the account is $2,040.40. To the nearest tenth of a percent, what is the annual interest rate for this account?

A deposit of $10,000 is placed in a savings account for 4 years. The interest for the account is compounded continuously. At the end of 4 years, the balance in the account is $12,068.34. What is the annual interest rate for this account. Round your answer to one decimal places.

An investment is made in an account that pays 9.25%, compounded daily. What is the effective yield for this investment? Round your answer to two decimal places.

An investment of $2500 is made in an account that compounds interest quarterly. After years 30 the balance in the account is $26132.78. To the nearest tenth of a percent, what is the annual interest rate for this account?

Solve the logarithmic equation. Round your answer to two decimal places. $\log _ { 2 } x + \log _ { 2 } ( x + 1 ) - \log _ { 2 } 3 = 4$

Find the principal that must be deposited in an account with an annual interest rate of 4%, compounded monthly, so that after 15 years the balance is $4,000. Round your answer to the nearest cent.

An investment of $500 is made in a trust fund at an annual interest rate of 5%, compounded monthly. How long will it take for the investment to double? Give your answer to two decimal places.

You deposit $2000 in an account. At the end of 21 years, your balance is $10671.45. The bank tells you that the annual interest rate for the account is 8%. How is the interest compounded?

Radioactive carbon, ${ } ^ { 14 } \mathrm { C }$ , has a has a half-life of 5730 years. After 1000 years, 3.7 grams remain. What was the initial quantity of ${ } ^ { 14 } \mathrm { C }$ ? Round your answer to two decimal places.

An investment of $28,000 is made in a trust fund at an annual interest rate of 6.75%, compounded continuously. How long will it take for the investment to double? Give your answer to two decimal places.

Solve the exponential equation $100,000 = 50,000 e ^ { 25 r }$ for r to determine the interest rate required for an investment of $50,000 to double in value when compounded continuously for 25 years. Round your answer to one decimal place.