How to Solve Exponential Growth and Decay Functions?
Exponential growth and decay are applied to physical quantities that rapidly change in value or form. Here you get familiarized with the concept of exponential growth and decay.
Exponential growth is when numbers increase exponentially, so there is a larger \(y\) for each \(x\)-value in the graph. Decay is when numbers decrease exponentially, so there is a smaller \(y\) for each \(x\)-value in the graph.
A step-by-step guide to exponential growth and decay
Exponential growth and decay apply to values that change rapidly. Exponential growth and decay are derived from the concept of geometric progression. Quantities that do not change as constant but change exponentially can be called exponential growth or exponential decay.
The simplest representation of exponential growth and decay is the formula \(ab^x\), where \(a\) is the initial quantity, \(b\) is the growth factor which is similar to the common ratio of the geometric progression, and \(x\) in the time steps for multiplying the growth factor. For exponential growth, the value of \(b\) is greater than \(1 (b>1)\), and for exponential decay, the value of \(b\) is lesser than \(1 (b< 1)\).
Exponential growth is used to study bacterial growth, population growth, and money growth schemes. Exponential decay refers to the rapid decrease of a value over some time.
The formulas of exponential growth and decay are presented below:
- Exponential growth:\(\color{blue}{f(x)=a(1+r)^t}\)
- Exponential decay:\(\color{blue}{f(x)=a(1- r)^t}\)
Exponential growth uses a factor \(r\) which is the rate of growth. The \(r\)-value lies between \(0\) and \(1\) \((0<r<1)\). The expression \((r + 1)\) can be considered a growth factor. And \(t\) is the time step which is the number of times the growth factor is to be multiplied. The value of \(t\) can be a whole number or a decimal number. For exponential decay, the growth factor is \((1 – r)\), which has a value less than \(1\).
Exponential Growth and Decay – Example 1:
If \($100,000\) is invested at a compound rate of \(4%\) per quarter, after \(2\) years, what is the amount received from the investment fund?
Solution:
The invested principal is \(a=$100,000\), the rate of compounding growth is \(r= 4%= 0.04\) per quarter.
The time is \(2\) years, and there are \(4\) quarters in a year, and we have \(t= 8\).
Using the concepts of exponential growth and decay, we have the following expressions for exponential growth:
\(f(x)=a(1 + r)^t\)
\(f(x)=100,000(1 + 0.04)^8\)
\(=100,000 (1.04)^8\)
\(=136856.90504\)
Therefore an amount of \($136,857\) is received after a period of \(2\) years.
Exercises for Exponential Growth and Decay
- The radioactive substance of thorium decomposes at a rate of \(7%\) per minute. What portion of \(15\) grams of thorium remains after \(6\) minutes?
- In \(2020\), \(200\) people lived in a remote town. The population has increased by \(20%\) every year. How many people will live in \(5\) years?
- \(\color{blue}{9.7}\)
- \(\color{blue}{498}\)
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