Question 1

If $18,000 is invested at an interest rate of 4.5% compounded continuously, what is the amount of the investment after 10 years? Rounded to the nearest hundredth.
A) $38,626.59
B) $38,040.78
C) $38,681.90
D) $18,000
E) $18,828.50

Accepted Answer

The formula for continuous compounding is $A = Pe^{rt}$, where $A$ is the amount of the investment after time $t$, $P$ is the principal amount ($18,000), $r$ is the annual interest rate (4.5% or 0.045), and $e$ is the base of the natural logarithm (approximately 2.71828). Plugging in the values, we get $A = 18000e^{0.045 \times 10}$. Calculating this gives us approximately $38,681.90, which matches choice C.

Question 2

If $16,000 is invested at an interest rate of 6% for 25 years, compounded continuously, determine the annual growth rate if the money were compounded quarterly? Rounded to the nearest thousandth of a percent.
A) 0.011%
B) 6.000%
C) 6.184%
D) 0.060%
E) 106.184%

Accepted Answer

We first use the formula $A=Pe^{rt}$ for continuous compounding, where $A$ is the ending amount, $P$ is the principal, $r$ is the annual interest rate, and $t$ is the time in years:
$$A=Pe^{rt}$$
$$A=16000e^{0.06\cdot 25}$$
$$A\approx 84967.88$$
Next, we use the formula $A=P\left(1+\frac{r}{n}ight)^{nt}$ for quarterly compounding, where $n$ is the number of times compounded per year and $t$ is the time in years:
$$84967.88=16000\left(1+\frac{r}{4}ight)^{4\cdot 25}$$
$$\frac{84967.88}{16000}=\left(1+\frac{r}{4}ight)^{100}$$
$$\ln\left(\frac{84967.88}{16000}ight)=100\ln\left(1+\frac{r}{4}ight)$$
$$\ln\left(\frac{84967.88}{16000}ight)=100\left(\frac{r}{4}-\frac{r^2}{32}+\frac{r^3}{96}-\cdotsight)$$
Since $r$ is small, we can approximate this as:
$$\ln\left(\frac{84967.88}{16000}ight)\approx 100\cdot \frac{r}{4}-100\cdot \frac{r^2}{32}$$
$$\frac{\ln\left(\frac{84967.88}{16000}ight)}{100}\approx \frac{r}{4}-\frac{r^2}{3200}$$
We can ignore the term $r^2/3200$ since it is much smaller than $r/4$ for small $r$, so we get:
$$\frac{\ln\left(\frac{84967.88}{16000}ight)}{100}\approx \frac{r}{4}$$
Solving for $r$, we get:
$$r\approx 0.06184\approx 6.184\%$$
Rounding to the nearest thousandth of a percent, we get:
$$\boxed{	extbf{(C) }6.184\%}$$

Question 3

On August 21, 2009, the World Health Organization announced its prediction that the number of new cases of H1N1 (swine flu) virus would double every 5 days for several months. As of July 27, 2009, the number of new cases was 16,409. Determine the number of new cases of the H1N1 virus after 9 days. Rounded to the nearest whole number.
A) 24,117
B) 57,139
C) 18,849
D) 8,401,408
E) 17,723

Accepted Answer

The answer of On August 21, 2009, the World Health...

Question 4

Solve the logarithmic equation  for  . Round to the nearest thousandth.
A) 10
B) 2
C) 1,024
D) 20
E) 100

Accepted Answer

The answer of Solve the logarithmic equation  for ...

Question 5

Solve the logarithmic equation  for  . Round to the nearest ten-thousandth.
A) 0.3333
B) 0.0833
C) -0.25
D) -1
E) -0.75

Accepted Answer

Applying logarithmic property

11ec1b6e_3c5a_aaf8_bbf8_496b9cac75ec_TBX8728_11 = 7 − log(11ec1b6e_3c5a_aaf9_bbf8_cb7e8736ca2a_TBX8728_11)

11ec1b6e_3c5a_aaf8_bbf8_496b9cac75ec_TBX8728_11 = log(10 / 120ecab8_6019_9f17_701b_aa7fe2eb9278_TBX8728_10)

11ec1b6e_3c5a_aaf8_bbf8_496b9cac75ec_TBX8728_11 = log(1 / 12)

11ec1b6e_3c5a_aaf8_bbf8_496b9cac75ec_TBX8728_11 = -log12

11ec1b6e_3c5a_aaf9_bbf8_cb7e8736ca2a_TBX8728_11 = 1/12

Therefore 11ec1b6e_3c5a_aaf9_bbf8_cb7e8736ca2a_TBX8728_11 ≈ 0.0833

Question 6

Solve the exponential equation  for  . Round to the nearest thousandth.
A) 0.604
B) 6.611
C) -0.274
D) 1.604
E) -0.230

Accepted Answer

The answer of Solve the exponential equation  for ...

Question 7

Solve the exponential equation  for  . Round to the nearest thousandth.
A) 0.217
B) 4.605
C) 273,252.194
D) 2.718
E) 2.000

Accepted Answer

Taking the natural log of both sides, we get:
ln(11ec1b6e_3c5a_35bc_bbf8_53c5a8ae3d8c_TBX8728_11) = ln(3) + ln(11ec1b6e_3c5a_35bc_bbf8_53c5a8ae3d8c_TBX8728_11) - ln(7)
Let x = ln(11ec1b6e_3c5a_35bc_bbf8_53c5a8ae3d8c_TBX8728_11). Then we have:
x = ln(3) + x - ln(7)
Solving for x:
x = ln(3/7)
Substituting back in for 11ec1b6e_3c5a_35bc_bbf8_53c5a8ae3d8c_TBX8728_11, we get:
11ec1b6e_3c5a_35bd_bbf8_cbb000a4d0a0_TBX8728_11 = e^(ln(3/7)) ≈ 0.4595
Rounding to the nearest thousandth, we get 0.459. Therefore, the answer is choice B.

Question 8

On August 21, 2009, the World Health Organization announced its prediction that the number of new cases of H1N1 (swine flu) virus would double every 3 days for several months. Determine the daily growth factor for the H1N1 virus (rounded to the nearest thousandths).
A) 8.000
B) 1.732
C) 2.000
D) 1.000
E) 1.260

Accepted Answer

The answer of On August 21, 2009, the World Health...

Question 9

Solve the exponential equation  for  . Round to the nearest thousandth.
A) 3.054
B) 0.602
C) 1.839
D) 2.882
E) 0.327

Accepted Answer

The answer of Solve the exponential equation  for ...

Question 10

On August 21, 2009, the World Health Organization announced its prediction that the number of new cases of H1N1 (swine flu) virus would double every 5 days for several months. As of July 27, 2009, the number of new cases was 26116. Which one of the following exponential functions model the growth of this virus?
A) 
B) 
C) 
D) 
E)

Accepted Answer

The exponential function for the growth of the H1N1 virus should have the form y = ae^(bx), where y is the number of cases, x is the time in days, and a and b are constants to be determined.

To find the values of a and b, we can use the given information that the number of new cases doubles every 5 days. This means that if we take the ratio of the number of cases at two different times 5 days apart, we should get a value of 2.

Let's use July 27th (day 0) and August 1st (day 5) as our first two data points. The number of new cases on July 27th was 26116. On August 1st, the number of new cases would be double that amount, or 52232.

Using the exponential function y = ae^(bx), and plugging in the values for y, x, and b (which is 1/5 since the virus doubles every 5 days), we get:

52232 = ae^(1/5 * 5)
52232 = ae

Simplifying this equation, we get:

a = 52232 / e

Now we have the value of a and can write the exponential function as:

y = (52232 / e) * e^(bx)

To find the value of b, we can use another set of data points. Let's use the number of new cases on July 27th and August 6th (day 10). The number of new cases on August 6th should be double the number on August 1st, or 104464.

Using the exponential function and plugging in the values, we get:

104464 = (52232 / e) * e^(1/5 * 10)

Simplifying this equation, we get:

e^(2/5) = 2

Taking the natural logarithm of both sides, we get:

ln(e^(2/5)) = ln(2)
2/5 = ln(2)
b = ln(2) / 5

Now we have the complete exponential function:

y = (52232 / e) * e^((ln(2) / 5)x)

Simplifying this equation, we get:

y = 52232 * 1.316^x

This equation corresponds to answer choice B.

Question 11

Solve the exponential equation  for  . Round to the nearest thousandth.
A) 1.946
B) 2.189
C) 12.772
D) -0.043
E) -23.133

Accepted Answer

The answer of Solve the exponential equation  for ...

Question 12

Using the graph below, determine which one of the following exponential functions models the graph. 
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Using the graph below, determine which one...

Question 13

Solve the logarithmic equation  for  .
A) 5
B) -5
C) 5.0
D) No solution
E) 0

Accepted Answer

The answer of Solve the logarithmic equation  for ...

Question 14

Solve the exponential equation  for  . Round to the nearest thousandth.
A) -0.924
B) 2.076
C) 2.303
D) 6.576
E) 4.000

Accepted Answer

The answer of Solve the exponential equation  for ...

Question 15

If $49,000 is invested at an interest rate of 2.5% per year, find the amount of the investment at the end of 20 years if the money is compounded monthly. Round to the nearest hundredth.
A) $50,239.13
B) $49,000
C) $80,292.21
D) $80,745.33
E) $51,082.58

Accepted Answer

We use the formula $A=P\left(1+\frac{r}{n}ight)^{nt}$ where $A$ is the amount at the end, $P$ is the principal, $r$ is the annual interest rate, $n$ is the number of times compounded per year, and $t$ is the number of years. Here, $P=49000$, $r=0.025$, $n=12$ (compounded monthly), and $t=20$. Substituting these values, we get $A=49000\left(1+\frac{0.025}{12}ight)^{12\cdot20}\approx80745.33$. Thus, the closest answer choice is $\boxed{	ext{(D) }80745.33}$.

Question 16

The number of bees in a certain area of Arizona is modeled by the function  where  is measured in number of weeks. Determine the number of bees after 17 weeks. Rounded to the nearest whole number.
A) 20,942.04
B) 5,822.67
C) 20,394.45
D) 16,022,649.18
E) 5,375

Accepted Answer

The answer of The number of bees in a certain...

Question 17

On August 21, 2009, the World Health Organization announced its prediction that the number of new cases of H1N1 (swine flu) virus would double every 5 days for several months. As of July 27, 2009, the number of new cases was 17,871. Determine the instantaneous growth rate for the virus (rounded to the nearest ten-thousandths).
A) 0.3495
B) 0.1386
C) 1.3979
D) 1.1487
E) 0.0602

Accepted Answer

The answer of On August 21, 2009, the World Health...

Question 18

Solve the logarithmic equation  for  . Round to the nearest hundredth.
A) 24.00
B) No Solution
C) 0.13
D) 0.38
E) -0.63

Accepted Answer

The answer of Solve the logarithmic equation  for ...

Question 19

Solve the exponential equation  for  . Round to the nearest thousandth.
A) 1.100
B) 1.230
C) 1.705
D) 0.813
E) 1.375

Accepted Answer

The answer of Solve the exponential equation  for ...

Question 20

Solve the logarithmic equation  for  . Round to the nearest hundredth.
A) 0.24
B) 0.15
C) 4.25
D) 0.29
E) 0.31

Accepted Answer

The answer of Solve the logarithmic equation  for ...

Quiz 4: Logarithmic Functions and Exponential Modela

If $18,000 is invested at an interest rate of 4.5% compounded continuously, what is the amount of the investment after 10 years? Rounded to the nearest hundredth.

If $16,000 is invested at an interest rate of 6% for 25 years, compounded continuously, determine the annual growth rate if the money were compounded quarterly? Rounded to the nearest thousandth of a percent.

Solve the logarithmic equation for . Round to the nearest thousandth.

Solve the logarithmic equation for . Round to the nearest ten-thousandth.

Solve the exponential equation for . Round to the nearest thousandth.

Solve the exponential equation for . Round to the nearest thousandth.

On August 21, 2009, the World Health Organization announced its prediction that the number of new cases of H1N1 (swine flu) virus would double every 3 days for several months. Determine the daily growth factor for the H1N1 virus (rounded to the nearest thousandths) .

Solve the exponential equation for . Round to the nearest thousandth.

Solve the exponential equation for . Round to the nearest thousandth.

Using the graph below, determine which one of the following exponential functions models the graph.

Solve the logarithmic equation for .

Solve the exponential equation for . Round to the nearest thousandth.

If $49,000 is invested at an interest rate of 2.5% per year, find the amount of the investment at the end of 20 years if the money is compounded monthly. Round to the nearest hundredth.

The number of bees in a certain area of Arizona is modeled by the function where is measured in number of weeks. Determine the number of bees after 17 weeks. Rounded to the nearest whole number.

Solve the logarithmic equation for . Round to the nearest hundredth.

Solve the exponential equation for . Round to the nearest thousandth.

Solve the logarithmic equation for . Round to the nearest hundredth.