Question

Rewrite the function in the form \(y=a(1+r)^t\) or \(y=a(1-r)^t\). Then state the growth or decay rate. \(y=a(0.25)^{t / 9}\)

Short answer

The function \(y = a(0.25)^{t / 9}\) can be rewritten in the form of \(y = a((1 - 0.75)^{(1/9)})^t\). It is a decay function with an annual decay rate of \(75\%\).

Step-by-step solution

Rewrite the Function

First, let's rewrite \(0.25\) as \(1 - 0.75\), which gives us \(y=a(1 - 0.75)^{t / 9}\). This is because \(1 - 0.75 = 0.25\). So our new function is \(y=a(1 - 0.75)^{t / 9}\). The \(0.75\) represents a decay rate because it is subtracted from 1.

Rewrite the Function to General Form

Following the common format of \(y=a(1-r)^t\), we could write this as \(y=a(1-0.75)^{t / 9} = a(0.25)^{t / 9}\). However, this isn't quite in the form we need because the exponent of the right side of the equation must be \(t\), not \(t / 9\). By utilising properties of exponents, we can adjust this to the desired form: \(t / 9 = t \times (1/9)\). This gives us \(y=a(1 - 0.75)^{t \times (1/9)}\). We know that \((a^b)^c = a^{b \times c}\), so this allows us to write this as \(y = a((1 - 0.75)^{(1/9)})^t\).

Identify the Decay Rate

The exponent of \(1/9\) inside the parenthesis can be considered as another form of rate, but in this case, it's an adjustment to the annual rate \(0.75\) over a 9-year period. Still, the main decay rate of the function is \(0.75\) or \(75\%\) annually.

Key concepts

Decay Rate

In the world of exponential decay, the decay rate is a crucial concept. It represents the percentage by which a quantity decreases over a specific period. This rate is often expressed as a decimal, where values less than one indicate decay.

In our example, we have the function: \[ y=a(1 - 0.75)^{t/9} \]Here, the decay rate is found in the term \(1 - r = 0.25\). Solving for \(r\) means recognizing that \(0.75\), which is subtracted from one, represents the decay rate. The \(0.75\) makes it clear that there is a 75% decrease occurring.

Thus, the decay rate refers to how much a function reduces or "decays." It is vital in various domains—like finance and physics—helping to predict outcomes based on exponential decline.

Remember:

• The decay rate is the amount subtracted from 1.
• In our example, it's \(0.75\) or 75%.
• This rate tells us how much the quantity decreases over a time period.

Exponential Function

An exponential function is a mathematical expression where a constant base is raised to a variable exponent. Such functions are powerful for modeling rapid changes in growth or decay.

Exponentials can take the form \(y = a(1 - r)^t\) or\(y = a(1 + r)^t\). Generally:

• The base (1-r) or (1+r) signifies decay and growth, respectively.
• The exponent \(t\) represents time.
• The initial value \(a\) dictates the starting quantity.

In our function \[ y=a(0.25)^{t/9} \],we observe that the change over time is both the base \(0.25\) and the time transformation \(t/9\). The key here is recognizing that the expression modifies over each time unit.

Exponential functions are a staple in fields analyzing growth and decay patterns, like population studies, radioactive decay, or investment portfolio performance.

Properties of Exponents

Understanding the properties of exponents is essential when handling exponential functions. These properties allow us to manipulate and simplify expressions such as \(y=a(1 - 0.75)^{t/9}\).

Some fundamental properties include:

• \((a^b)^c = a^{b\times c}\) - This shows how we can deal with nested exponents.
• \(a^{b/c} = \sqrt[c]{a^b}\) - Root transformations that apply to our expression, especially when time periods are intervalled.
• \(a^0 = 1\) - Any base raised to the power zero equals one.

In our function, changing \(t/9\) into an expression involving \(t\) is done through this property. Here's how:
First,\[(1-r)^{t/9} = ((1-r)^{1/9})^t\]By applying\((a^b)^c = a^{b \times c}\), we successfully align with the exponential function's required form.

Being adept with these rules ensures smooth handling of complex exponential functions tasks, providing clarity and accuracy in problem-solving.