Category Archives: 5.5 Applications of Exponential and Logarithmic Functions

5.5 Applications of Exponential and Logarithmic Functions, 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models, Applications of Exponential Functions

Application of Exponential Functions: Exponential Decay (Estimate Percentage Remaining given the Half-life)

July 27, 2020 admin

5.5 Applications of Exponential and Logarithmic Functions, 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models, Applications of Exponential Functions

Application of Exponential Functions: Population Growth

July 5, 2020 admin

5.5 Applications of Exponential and Logarithmic Functions, 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

Application: Exponential Growth

July 20, 2018 admin

5.5 Applications of Exponential and Logarithmic Functions, 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

Application: Exponential Growth and Decay

July 20, 2018 admin

5.5 Applications of Exponential and Logarithmic Functions, 6.7 Financial Models, Applications of Exponential Functions

Application of Exponential Functions: Doubling Time

July 11, 2018 admin

Example:

How long does it take for an investment to double if it is invested at 18% compounded continuously?

Solution:

Since this question involve continuous compound interest, we will use the associated formula.

$A=Pe^{rt}$

We are given that the interest rate is 18% or 0.18. This tells me that when r=0.18 Since we are looking for the doubling time, A will be 2 times P. I can write that in symbols A=2P.

Substitute these values into the continuous compound formula and solve for the interest rate.

Continuous compound formula	$A=Pe^{rt}$
Substitute the values of r and A into the formula	$A=Pe^{rt}$ $2P=Pe^{0.18*t}$ $2P=Pe^{0.18t}$
Solve for t by dividing both sides by P and simplifying	$2P=Pe^{0.18t}$ ${2P}/P={Pe^{0.18t}}/P$ $2=e^{0.18t}$
Solve for t by taking the log of both sides.	$2=e^{0.18t}$ $ln 2=ln e^{0.18t}$
Solve for t by using the power rule and simplifying	$ln 2=ln e^{0.18t}$
Solve for t by dividing both sides by 0.18 and simplifying
Find the value in the calculator
Write the answer rounded to two decimal places

It will take 3.85 years to double your money when interest is compounded continuously at 18%.

If you need to write this in years and months, you will need to convert the 0.85 to months. Since there are 12 months in a year, multiply 0.85 by 12 to get 10.2. I will round to the nearest months to get 10.

It will take 3 years and 10 months to double your money when interest is compounded continuously at 18%.

Here is a video that is similar except that you are looking for the investment to triple.

5.5 Applications of Exponential and Logarithmic Functions, 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

Application: Exponential Growth and Decay (Half-life)

October 28, 2017 admin

Example: The half life of radium is 1690 years. If 50 grams are present now, how much will be present in 630 years?

Solution: There is a two part process to this problem. Part 1: Use some of the information to find the decay rate of radium. Part 2: Answer the question using the rest of the given information.

Part 1: Find the decay rate of radium.

Since we are using an exponential model for this problem we should be clear on the parts of the exponential decay model.

Exponential Decay Model

$A=A _0 e^{kt}$

A _0 is the initial amount

is the decay rate

is the time

is the amount after t time has passed

Since radium has a half life of 1690 years, we know that after 1690 years there will be half of the initial amount of radium left. This allows me to establish a relationship between the initial amount and the amount after 1690. The amount after 1690 year is half of the initial amount. A=1 /2 A _0 when the t=1690 .

Substitute these values into the exponential decay formula and solve for k.

Exponential Decay Formula	$A=A _0 e^{kt}$
Make a substitution for A and t since it is known that the half-life is 1690 years and	$A=A _0 e^{kt}$ $1 /2 A _0=A _0 e^{k(1690)}$
Solve for the decay rate k:Start by dividing both sides by the coefficient to isolate the exponential factor	$1 /2 A _0=A _0 e^{k(1690)}$ ${1 /2 A _0}/A_0={A _0 e^{k(1690)}}/A_0$ $1 /2= e^{k(1690)}$
Solve for the decay rate k:Take the natural log of both sides to get k out of the exponent	$1 /2= e^{k(1690)}$ $ln(1 /2)= ln(e^{k(1690)})$
Solve for the decay rate k:Use the power rule for logarithms to get k out of the exponent	$ln(1 /2)= ln(e^{k(1690)})$
Solve for the decay rate k:Simplify ln e = 1
Solve for the decay rate k:Solve for k by dividing by 1690 on both sides

Using only the information about radium having a half-life of 1690 years I have found the decay rate for radium.

k={ln(1 /2)}/1690
k approx -0.000410

Note: Although I have put an approximation for k here, try not to round until the very last step.

Part 2: Answer the question using the rest of the given information.

Given information: If 50 grams are present now, how much will be present in 630 years?

With this information I can identify the initial amount of radium as 50 grams and the time to be 630 years. Symbolically that is A _0=50 when the t=630 . Substitute the given information and the decay rate k found in part 1 to the exponential decay formula.

Exponential Decay Formula	$A=A _0 e^{kt}$
Make a substitution for initial amount , time (t), and the decay rate (k). , , and	$A=A _0 e^{kt}$ $A=50 e^{630{ln(1 /2)}/1690}$
Type the expression into your calculator and round to the thousandth place.	$A=50 e^{630{ln(1 /2)}/1690}$

A approx 38.615

38.615 grams will be present 630 years later is 50 grams are present initially.

Here is a video example that is similar to the above example and it shows how to enter information in the calculator.

5.5 Applications of Exponential and Logarithmic Functions, 6.7 Financial Models

Application of Exponential Functions: Finding the Interest Rate

August 4, 2017 admin

Example:

What is the interest rate necessary for an investment to quadruple after 7 year of continuous compound interest?

Solution:

Since this question involve continuous compound interest, we will use the associated formula.

$A=Pe^{rt}$

We are given that the invest quadruples in 7 years. This tells me that when t=7 that A will be 4 times P. I can write that in symbols A=4P.

Substitute these values into the continuous compound formula and solve for the interest rate.

Continuous compound formula	$A=Pe^{rt}$
Substitute the values of t and A into the formula	$A=Pe^{rt}$ $4P=Pe^{r*7}$ $4P=Pe^{7r}$
Solve for r by dividing both sides by P and simplifying	$4P=Pe^{7r}$ ${4P}/P={Pe^{7r}}/P$ $4=e^{7r}$
Solve for r by taking the log of both sides.	$4=e^{7r}$ $ln 4=ln e^{7r}$
Solve for r by using the power rule and simplifying	$ln 4=ln e^{7r}$
Solve for r by dividing both sides by 7 and simplifying
Find the value in the calculator
Write the answer as a percentage rounded to two decimal places