Problem 52 - College Algebra

Question

For the following exercises, evaluate the natural logarithmic expression without using a calculator. $$\ln \left(e^{-0.225}\right)-3$$

Step-by-Step Solution

Answer

The expression evaluates to \(-3.225\).

Step 1: Understand the Problem

We are given the natural logarithmic expression \( \ln(e^{-0.225}) - 3 \) and need to evaluate it without using a calculator. The expression involves the natural log function and the exponential function.

Step 2: Simplify the Logarithmic Expression

Recall the property of logarithms: \( \ln(e^x) = x \). Apply this property to the expression \( \ln(e^{-0.225}) \), which simplifies to \( -0.225 \).

Step 3: Evaluate the Expression

Substitute the simplified result back into the original expression: \(-0.225 - 3\).

Step 4: Perform the Arithmetic Operation

Subtract 3 from \(-0.225\):\[-0.225 - 3 = -3.225\].

Step 5: Verify the Result

Review each step to ensure correctness. We've applied the logarithmic identity correctly and performed simple arithmetic accurately.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. They take the form \(a^x\), where \(a\) is a positive constant, and \(x\) is the exponent. A common example is the natural exponential function, where the base \(a\) is Euler's number \(e\), approximately 2.71828. This is expressed as \(e^x\).
Exponential functions are widely used because they exhibit rapid growth or decay, depending on whether the exponent is positive or negative.

• For positive \(x\), \(e^x\) increases quickly.
• For negative \(x\), \(e^x\) decreases, approaching zero.

In the context of our example, we focus on \(e^{-0.225}\), an exponential decay situation where the exponent is negative. The properties of such functions are central to simplifying expressions involving the natural logarithm, as they provide a straightforward relationship captured by logarithmic properties.

Logarithmic Properties

Logarithmic properties help us solve problems involving logarithms by offering specific rules. These rules include the natural logarithm, denoted \(\ln\), with base \(e\). A key property is \(\ln(e^x) = x\), which states that the natural logarithm of \(e\) raised to any power \(x\) equals that power. In essence, it illustrates the inverse relationship between logarithms and exponentials.
Utilizing this property, calculating \(\ln(e^{-0.225})\) simplifies to \(-0.225\), as it's essentially just the exponent in question. This transformation is crucial to making problems involving logarithms more approachable.

• Recalling and applying this property quickly reduces complexity in expressions.
• Helps in decoding the relationship between exponential growth/decay and logarithmic scales.

This property is a guiding light in moving from a complex-looking expression back to a simple numerical value.

Arithmetic Operations

Arithmetic operations involve basic mathematical procedures like addition, subtraction, multiplication, and division. In the context of the given expression \(-0.225 - 3\), we specifically engage in subtraction. Subtraction helps determine the difference between numbers, and the task is straightforward: deduct 3 from \(-0.225\).
Here's a simplified step to perform the operation:

• Rewrite \(-0.225 - 3\) as adding a negative: \(-0.225 + (-3)\).
• Think of this as going \(-3.225\), since adding negatives is akin to subtracting positives.

This operation results in the final answer \(-3.225\). Understanding such operations is vital because they allow you to accurately navigate through relationships between numbers, aiding in the comprehensive resolution of mathematical expressions.