Math Problem Statement

A quantity with an initial value of 990 grows exponentially at a rate such that the quantity doubles every 4 years. What is the value of the quantity after 0.7 decades, to the nearest hundredth?

Solution

The problem describes an exponential growth situation where the quantity doubles every 4 years. Let's model this situation using the exponential growth formula:

Q(t)=Q0ektQ(t) = Q_0 \cdot e^{kt}

Where:

  • Q(t)Q(t) is the quantity at time tt,
  • Q0Q_0 is the initial quantity,
  • kk is the growth rate constant,
  • tt is time.

Step 1: Use the doubling time to find kk

We are told that the quantity doubles every 4 years, meaning that:

Q(4)=2Q0Q(4) = 2 \cdot Q_0

Substitute into the exponential growth formula:

2Q0=Q0ek42 \cdot Q_0 = Q_0 \cdot e^{k \cdot 4}

Cancel Q0Q_0 (since it's nonzero):

2=e4k2 = e^{4k}

Now, solve for kk:

ln(2)=4k\ln(2) = 4k

k=ln(2)40.693140.1733k = \frac{\ln(2)}{4} \approx \frac{0.6931}{4} \approx 0.1733

Step 2: Find the value of the quantity after 0.7 decades

Now, we want to find the value of the quantity after 0.7 decades. Since 1 decade equals 10 years, 0.7 decades is 0.7×10=70.7 \times 10 = 7 years.

Substitute k0.1733k \approx 0.1733 and t=7t = 7 years into the formula:

Q(7)=990e0.17337Q(7) = 990 \cdot e^{0.1733 \cdot 7}

Calculate the exponent:

0.17337=1.21310.1733 \cdot 7 = 1.2131

Now, calculate Q(7)Q(7):

Q(7)=990e1.21319903.364Q(7) = 990 \cdot e^{1.2131} \approx 990 \cdot 3.364

Q(7)3321.36Q(7) \approx 3321.36

Final Answer:

The value of the quantity after 0.7 decades is approximately 3321.36 to the nearest hundredth.


Would you like further details or clarifications on any step?

Here are 5 related questions to consider:

  1. How do you convert exponential growth formulas for different time intervals?
  2. How does the base of the natural logarithm influence the growth rate?
  3. What happens if the doubling time changes?
  4. How can we calculate the decay rate if the quantity is shrinking instead of growing?
  5. How would the growth rate change if the time to double the quantity were halved?

Tip: When dealing with exponential growth or decay problems, always remember that the rate constant kk can be found using the doubling or halving time formula, as shown in the problem!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Exponential Growth
Natural Logarithms
Growth Rate
Time Calculations

Formulas

Q(t) = Q_0 * e^(kt)
Q(4) = 2 * Q_0
ln(2) = 4k
k = ln(2)/4

Theorems

Exponential Growth Formula
Doubling Time Formula

Suitable Grade Level

Grades 11-12