Problem 60 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, solve for \(x\) or \(t\). $$ e^{-0.5 x}=0.075 $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 5.1806\).

Step 1: Apply the natural logarithm to both sides

To remove the base \(e\) from the left-hand-side, apply the natural logarithm \(ln\) to both sides of the given equation. The natural logarithm of a number \(b\), denoted \(ln(b)\), is the power to which \(e\) would have to be raised to equal \(b\). So, applying natural log to both sides, the equation becomes: \[ ln(e^{-0.5 x}) = ln(0.075)\]

Step 2: Use logarithmic identity on the left-hand-side

The natural logarithm of the exponential function is just the function itself. Thereby, \(-0.5x\) is simply left on the left-hand-side: \[-0.5x = ln(0.075)\] The left-hand-side can be rearranged by multiplying by \( -2\), to isolate the variable on the left-hand-side. This results to: \[ x = -2 \times ln(0.075) \].

Step 3: Evaluate the expression

Finally, calculate the value of the right-hand side using a calculator or any computing tool (like Python, Excel, etc.) that allows the computation of the natural logarithm: \[ x = -2 \times -2.5903.\]

Key Concepts

Exponential Equation

When working with equations, an **exponential equation** is one that sets a constant base raised to a variable in equal standing with a constant or another variable expression.
Exponential equations often involve the base number of mathematics, the natural number \( e \), which is approximately 2.718.
Here, our base \( e \) is raised to the power of \(-0.5x\), with both sides of the equation having to balance: \( e^{-0.5x} = 0.075 \).

• **Recognize the Base:** The base \( e \) is common in natural growth scenarios and certain mathematical models.
• **Understanding Exponents:** An exponent indicates how many times to multiply the base by itself. Here, we have a negative exponent, which often indicates a division or reciprocal is involved.
• **Solving Approach:** It is critical to isolate the exponents when solving such equations by using logarithms, as exponential growth occurs when a calculation repeatedly applies multiplication.

By taking these steps, we transform the equation and use other mathematical strategies to find the solution for the unknown variable.

Solving for a Variable

To **solve for a variable** like \( x \) in an equation, you must effectively isolate the variable on one side of the equation, making its value apparent.
In the given problem, the equation \( e^{-0.5x} = 0.075 \) requires us to remove the exponentiation aspect to help isolate \( x \).

• **Applying the Natural Logarithm:** Taking the natural logarithm on both sides allows you to convert the exponential form to a more straightforward expression.
• **Using the Identity:** The logarithmic identity simplifies the exponent, eliminating the exponential and revealing the multiplication aspect, namely \(-0.5x\).
• **Manipulating the Equation:** By dividing or multiplying all terms, given the coefficient \(-0.5\), you can isolate and solve for the variable \( x \).

This methodical approach eases understanding how the variable relates to known quantities, facilitating solution calculations.

Logarithmic Identity

A **logarithmic identity** is a mathematical expression that helps simplify equations involving logarithms.
In the given exercise, we utilize the fact that the natural logarithm \( ln \) of an exponential function with base \( e \) will revert to its exponent.

• **Understanding the Identity:** ln(e^y) = y, which indicates applying the natural log to any base \( e \) exponent returns the exponent itself, simplifying the equation.
• **Application in the Example:** This identity directly translates \( ln(e^{-0.5x}) \) to \( -0.5x \), bypassing the complexity of the exponent.
• **Benefits of the Identity:** It simplifies the steps needed to solve and balances equations that involve natural logarithms instantly.

Using such identities efficiently is crucial for solving exponential and logarithmic equations, as they reduce potential algebraic errors and streamline calculations.