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:
Where:
- is the quantity at time ,
- is the initial quantity,
- is the growth rate constant,
- is time.
Step 1: Use the doubling time to find
We are told that the quantity doubles every 4 years, meaning that:
Substitute into the exponential growth formula:
Cancel (since it's nonzero):
Now, solve for :
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 years.
Substitute and years into the formula:
Calculate the exponent:
Now, calculate :
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:
- How do you convert exponential growth formulas for different time intervals?
- How does the base of the natural logarithm influence the growth rate?
- What happens if the doubling time changes?
- How can we calculate the decay rate if the quantity is shrinking instead of growing?
- 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 can be found using the doubling or halving time formula, as shown in the problem!
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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
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