Problem 10 - Applied Calculus

Question

For Problems \(1-16,\) solve for \(t\) using natural logarithms. $$10=6 e^{0.5 t}$$

Step-by-Step Solution

Answer

\( t = \frac{\ln\left(\frac{5}{3}\right)}{0.5} \)

Step 1: Isolate the Exponential Term

First, we need to isolate the exponential part of the equation. The given equation is \( 10 = 6e^{0.5t} \). To isolate \( e^{0.5t} \), divide both sides by 6: \( \frac{10}{6} = e^{0.5t} \). Simplifying \( \frac{10}{6} \) gives \( \frac{5}{3} \), so the equation becomes \( \frac{5}{3} = e^{0.5t} \).

Step 2: Apply Natural Logarithm

Next, we apply the natural logarithm to both sides of the equation to eliminate the exponential. So, take the natural logarithm of both sides: \( \ln\left(\frac{5}{3}\right) = \ln\left(e^{0.5t}\right) \). Using the property of logarithms that \( \ln(e^x) = x \), we simplify the right side to \( 0.5t \), resulting in \( \ln\left(\frac{5}{3}\right) = 0.5t \).

Step 3: Solve for t

To solve for \( t \), we divide both sides of the equation by 0.5: \( t = \frac{\ln\left(\frac{5}{3}\right)}{0.5} \). This equation gives the value of \( t \) in terms of natural logarithms.

Key Concepts

Solving Equations

Solving equations is a fundamental skill in mathematics that involves finding the value of a variable that makes an equation true. In our original exercise, we need to solve the equation for the variable \( t \):

• The goal is to transform the equation step by step until \( t \) is isolated on one side.
• This process was demonstrated by first isolating the exponential term and then by applying logarithmic operations.

Keep in mind that solving any equation generally requires balancing both sides of the equation. To maintain equality, whatever operation you apply to one side, you must do to the other. This ensures that the equation remains valid through your transformations.
By understanding these steps, you can solve a wide range of mathematical problems.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent, like \( e^{0.5t} \). They are widespread in modeling naturally occurring processes such as population growth or radioactive decay.
In our exercise, the function \( e^{0.5t} \) shows how the variable \( t \) influences the overall value:

• Initially, the equation was in the form \( 10 = 6e^{0.5t} \).
• By isolating the exponential term, we can focus on its impact within the equation.

Such functions grow or diminish exponentially, meaning even small changes in \( t \) can cause significant variations in the output. This nature of exponential growth and decay forms the basis of why these functions are incredibly powerful in both theoretical and applied mathematics.

Logarithmic Properties

Logarithmic properties are essential tools when working with exponential functions. The natural logarithm, denoted as \( \ln \), helps us "undo" exponentials, providing an easy way to solve equations involving exponential terms.

• In the solution, the natural logarithm is applied to both sides of the equation \( \frac{5}{3} = e^{0.5t} \).
• This utilizes the logarithmic property that \( \ln(e^x) = x \).

By applying \( \ln \), we can simplify the equation by removing the exponential, making \( 0.5t \) directly accessible. Remember:

• Logarithms are the inverse operations of exponentiation, much like division is to multiplication.
• Understanding these properties allows for simplification and isolation of variables within exponential equations.

Logarithmic properties transform complicated expressions into more manageable forms, aiding in both theoretical insights and practical calculations.