Problem 67 - Intermediate Algebra

Question

Solve each equation. Express all answers to four decimal places. $$ \ln x=1.001 $$

Step-by-Step Solution

Answer

The solution for \( \ln x = 1.001 \) is approximately \( x = 2.7211 \).

Step 1: Understand the Equation

The equation given is \( \ln x = 1.001 \). The function \( \ln x \) represents the natural logarithm of \( x \). We need to find the value of \( x \) such that its natural logarithm is equal to 1.001.

Step 2: Exponentiate Both Sides

To isolate \( x \), exponentiate both sides of the equation using the base of the natural logarithm, \( e \). This gives us \( x = e^{1.001} \).

Step 3: Calculate \( e^{1.001} \)

Use a calculator to compute the value of \( e^{1.001} \). This calculation will give us the numerical value of \( x \).

Step 4: Round to Four Decimal Places

After calculating, we find \( e^{1.001} \approx 2.7210 \). Round this value to four decimal places, which results in the answer: \( x \approx 2.7211 \).

Key Concepts

Exponentiation

Exponentiation is a mathematical operation in which a number called the base is raised to the power of an exponent. In simpler terms, it means multiplying a number by itself a certain number of times. The operation is denoted as \( b^n \), where \( b \) is the base and \( n \) is the exponent.

• For example, \( 3^2 \) means \( 3 \times 3 \), which equals \( 9 \).
• If the exponent is \( 1 \), the base remains unchanged (e.g., \( 4^1 = 4 \)).
• Raising a base to the power of \( 0 \) always results in \( 1 \), regardless of the base (e.g., \( 5^0 = 1 \)).

In the context of natural logarithms, the base of the logarithm is \( e \), a mathematical constant approximately equal to \( 2.71828 \). Thus, when we encounter an equation like \( \ln x = 1.001 \), we use the exponentiation operation on both sides to solve for \( x \):
\[ x = e^{1.001} \]Here, we are essentially "undoing" the logarithm to reveal \( x \). This approach is key in solving equations involving natural logarithms.

Numerical Calculation

Numerical calculations involve the precise computation of mathematical expressions using numbers, calculators, or computational software. When solving problems like finding \( e^{1.001} \), accuracy is essential to achieve a reliable result.
Here's how to perform the calculation:

• Use a scientific calculator to compute expressions involving mathematical constants like \( e \). Many calculators have a specific button for \( e \) and its exponentiation.
• Enter the value \( 1.001 \) when prompted for the exponent.
• Press the button designated for exponentiation with base \( e \) to obtain the result. In this case, we find \( e^{1.001} \approx 2.7210 \).

For such calculations, it's important to ensure your calculator is functioning correctly and set to the appropriate mode (e.g., degree/radian mode for trigonometric functions, although not needed here). This vigilance ensures the accuracy of numerical results.

Rounding Decimals

Rounding decimals is a technique used to simplify numbers by reducing digits after the decimal point, making them easier to manage while maintaining a reasonable approximation to the original number.
Here's a brief guide to rounding:

• Identify the number of decimal places required. In our problem, we need four decimal places.
• Look at the digit immediately following the last required decimal place. If it's \( 5 \) or greater, round up the last kept digit by \( 1 \). If it's less than \( 5 \), leave the last kept digit unchanged.
• Apply this logic: For \( 2.7210 \) to be rounded to four decimal places, look at the fifth decimal (if available). Since that digit is \( 0 \), we do nothing and keep it as \( 2.7210 \).

Rounding ensures numbers are precise enough for practical use yet simplified to avoid unnecessary complexity. This makes data more approachable and calculations more efficient. In mathematical contexts, especially exams or standardized tests, pay attention to rounding instructions found in the problem statement.