Problem 7
Question
Identify the initial amount and the growth rate in the exponential function. $$y=100(1+0.5)^{t}$$
Step-by-Step Solution
Verified Answer
The initial amount is 100 and the growth rate is 0.5 in the exponential function \(y=100(1+0.5)^{t}\).
1Step 1: Identification of initial amount
In the given function \(y=100(1+0.5)^{t}\), the number 100 is the coefficient of the exponential term, which corresponds to \(a\) in the formula. Hence, the initial amount, \(a\), is 100.
2Step 2: Identification of growth rate
Looking at the part inside the parentheses in the function, the calculation (1+0.5) is observed. The 0.5 corresponds to \(r\) in the formula, indicating the growth rate. Since the 0.5 is added to 1, it means the growth rate is positive, thus indicating a growth process. Hence, the growth rate, \(r\), is 0.5.
Key Concepts
Initial Amount in Exponential FunctionsGrowth Rate in Exponential FunctionsExponential GrowthAlgebraic Problem Solving
Initial Amount in Exponential Functions
Understanding the initial amount in an exponential function is crucial for comprehending how these functions model various real-world processes, such as population growth or radioactive decay. In the equation \(y = a(1 + r)^t\), the initial amount is represented by \(a\). This is the value of the function when \(t = 0\), the starting point before any growth or decay has occurred.
In the sample exercise, where we have the function \(y = 100(1 + 0.5)^t\), the initial amount is simply 100. This could represent a starting population of 100 organisms, or an initial investment of $100, for example. When solving similar problems, identifying this value is the first step to understanding the function's behavior over time.
In the sample exercise, where we have the function \(y = 100(1 + 0.5)^t\), the initial amount is simply 100. This could represent a starting population of 100 organisms, or an initial investment of $100, for example. When solving similar problems, identifying this value is the first step to understanding the function's behavior over time.
Growth Rate in Exponential Functions
The growth rate in an exponential function dictates how quickly the quantity increases over time. It is denoted by \(r\) in the general form \(y = a(1 + r)^t\), where \(r\) is expressed as a decimal value. A positive rate signifies growth, while a negative rate indicates decay.
In the function \(y = 100(1 + 0.5)^t\), the growth rate is 0.5 or 50%. This high rate suggests that the quantity is doubling in shorter time intervals. Understanding the growth rate allows us to predict the future values of the function, critical for planning, and forecasting in various fields such as finance and biology.
In the function \(y = 100(1 + 0.5)^t\), the growth rate is 0.5 or 50%. This high rate suggests that the quantity is doubling in shorter time intervals. Understanding the growth rate allows us to predict the future values of the function, critical for planning, and forecasting in various fields such as finance and biology.
Exponential Growth
Exponential growth occurs when a quantity increases at a rate proportional to its current value, leading to faster and faster growth as time passes. It's exemplified by the general equation \(y = a(1 + r)^t\), where each time period sees the quantity multiplied by a constant factor.
In scenarios like compounding interest or unchecked population growth, this concept is vital. Back to our exercise, with a 50% growth rate, if 100 units grow exponentially, after one time period, there would be 150 units, then 225 in the next, and so on, illustrating the rapid increase. It's important for students to recognize the potential implications of exponential growth in practical scenarios.
In scenarios like compounding interest or unchecked population growth, this concept is vital. Back to our exercise, with a 50% growth rate, if 100 units grow exponentially, after one time period, there would be 150 units, then 225 in the next, and so on, illustrating the rapid increase. It's important for students to recognize the potential implications of exponential growth in practical scenarios.
Algebraic Problem Solving
Algebraic problem solving involves identifying variables and constants, understanding formulas, and manipulating these to find solutions. It's a step-by-step process that requires careful analysis and strategic application of algebraic rules.
In the context of exponential functions, problem-solving may require isolating the variable, using logarithms to find unknown exponents, or substituting known values to determine unknown quantities. Improving these skills can help students approach problems systematically, critical for their success in mathematics and numerous applications that require analytical thinking.
In the context of exponential functions, problem-solving may require isolating the variable, using logarithms to find unknown exponents, or substituting known values to determine unknown quantities. Improving these skills can help students approach problems systematically, critical for their success in mathematics and numerous applications that require analytical thinking.
Other exercises in this chapter
Problem 6
Use the product of powers property to write the expression as a single power of the base. \(a^{4} \cdot a^{6}\)
View solution Problem 7
Evaluate the expression without using a calculator. $$ 2^{-4} \cdot 2^{5} $$
View solution Problem 7
Write the number in decimal form. $$ 2.45 \times 10^{-1} $$
View solution Problem 7
Use the quotient of powers property to simplify the expression. $$ \frac{x^{12}}{x^{9}} $$
View solution