Problem 27 - Applied Calculus

Question

Put the functions in the form \(P=P_{0} e^{k t}\). $$P=15(1.5)^{t}$$

Step-by-Step Solution

Answer

Rewritten form: \(P = 15 e^{0.4055t}\)

Step 1: Identify the initial form

The given function is \( P = 15(1.5)^t \). To rewrite it in the form \( P = P_0 e^{kt} \), we need to find appropriate values for \( P_0 \) and \( k \).

Step 2: Identify \(P_0\)

In the function \( P = 15(1.5)^t \), \( P_0 \) corresponds to the coefficient of the exponential base, which is 15. So, \( P_0 = 15 \).

Step 3: Convert base to exponent with base e

Now, we need to express \( 1.5^t \) in the form of an exponential function with base \( e \). Recall that \( a^x = e^{x \ln a} \). Thus, \( 1.5^t = e^{t \ln 1.5} \).

Step 4: Determine k

Using the conversion \( 1.5^t = e^{t \ln 1.5} \), we identify that \( k = \ln 1.5 \). Calculate \( k \) using logarithms: \( k \approx 0.4055 \) (since \( \ln 1.5 \approx 0.4055 \)).

Step 5: Write the transformed equation

Substitute \( P_0 = 15 \) and \( k = \ln 1.5 \) into \( P = P_0 e^{kt} \), resulting in \( P = 15 e^{0.4055 t} \).

Key Concepts

Mathematical Transformation

The process of mathematical transformation allows us to convert equations into different forms without changing their meaning. Doing so can make them easier to understand or more convenient for further calculations. In the given exercise, we focused on transforming the function from one form to another to leverage the natural exponential base.

For example, the original function was given in the form of \( P = 15(1.5)^t \). The task was to manipulate this into a form that uses the base \( e \), which is common for exponential functions in calculus and advanced mathematics. Ensuring the transformation retains the original relationship is crucial, so it's important to precisely calculate each element during the conversion.

Such transformations are widespread in mathematical applications, as different problems might require the equation to be presented in particular formats for simplifying integration or differentiation, which often involves the natural exponential base \( e \).

Base e Conversion

Converting exponential functions to have a base \( e \) is valuable for simplifying many mathematical operations. The constant \( e \) is approximately equal to 2.71828 and is the base of natural logarithms, which makes it highly significant in mathematics.

When working with the function \( P = 15(1.5)^t \), the key step was to change the base from 1.5 to \( e \). This is possible by using the relationship \( a^x = e^{x \ln a} \). This formula is used to express any base \( a \) as an exponent of \( e \).

• Identify the original base, which is 1.5 in this case.
• Apply the transformation formula to rewrite it: \( 1.5^t = e^{t \ln 1.5} \).
• This changes our equation into a form where the base is \( e \), making it \( P = 15 e^{t \ln 1.5} \).

Performing a base conversion simplifies exponential functions, especially when dealing with calculus applications, as derivatives and integrals of exponential functions with base \( e \) are more straightforward.

Natural Logarithms

Natural logarithms are logarithms with base \( e \). They play a critical role in mathematics due to their relationship with exponential functions. When we talk about the natural logarithm of a number, it is expressed as \( \ln(x) \), and it essentially answers the question: 'To what power must \( e \) be raised to yield \( x \)?'.

In the transformation of \( P = 15(1.5)^t \) to a natural exponential form, natural logarithms helped determine the value of \( k \). By using the fact that \( 1.5^t = e^{t \ln 1.5} \), it was identified that \( k = \ln 1.5 \).

• Calculate \( k \) with the natural logarithm: \( k = \ln 1.5 \), which approximately equals 0.4055.
• Substitute \( k \) back into the equation to conclude the transformation.

Natural logarithms are indispensable when working with exponential functions, as they simplify the process of solving equations and understanding growth patterns, thanks to their nice properties when differentiating or integrating exponential expressions.