Problem 49 - Algebra 2

Question

Solve each equation or inequality. Round to the nearest ten-thousandth. \(\ln (x+1)=1\)

Step-by-Step Solution

Answer

x = 1.7183

Step 1: Isolate the Exponential Expression

The given equation is \(\ln(x+1) = 1\). To eliminate the natural logarithm, we must exponentiate both sides of the equation using base \(e\). This means applying \(e^\) to both sides: \(e^{\ln(x+1)} = e^1\).

Step 2: Simplify the Equation

Exponentiating cancels out the natural logarithm on the left side, so we end up with \(x+1 = e\).

Step 3: Solve for x

To solve for \(x\), we need to isolate it by subtracting 1 from both sides of the equation: \(x = e - 1\).

Step 4: Calculate e and Simplify

Evaluate the expression \(e - 1\). The approximate value of \(e\) is 2.7183. Therefore, \(x = 2.7183 - 1 = 1.7183\).

Step 5: Round Solution

Round the result to the nearest ten-thousandth. The value of \(x\) rounded to the nearest ten-thousandth is \(x = 1.7183\).

Key Concepts

Understanding Natural Logarithms

Natural logarithms, symbolized by \( \ln \), are logarithms with the base \( e \). The number \( e \) is a fundamental constant approximately equal to 2.7183. Using the natural logarithm allows mathematicians and scientists to simplify calculations involving exponential growth or decay.
For example, in the equation \( \ln(x+1) = 1 \), the natural logarithm indicates the power to which \( e \) must be raised to give the value \( x+1 \). This expression can be transformed by applying the exponential function \( e \) to both sides. Doing so effectively "undoes" the logarithm, allowing us to proceed with solving the equation. Understanding how to manipulate natural logarithms is a key skill in math, enabling you to solve complex problems in both academics and real-world applications.

Solving Exponential Equations

Solving exponential equations involves isolating the exponential term. When you see an equation featuring a logarithm like \( \ln(x+1) = 1 \), the first step is to remove the logarithmic function. By exponentiating both sides of the equation using the base \( e \), you can cancel out the natural logarithm:
- \( e^{\ln(x+1)} = e^1 \) becomes \( x+1 = e \).
This transformation simplifies the equation, allowing you to move to the next step: solving for the unknown variable \( x \).
- Subtract 1 from both sides to isolate \( x \), resulting in \( x = e - 1 \).
Exponential equations are common in various fields such as finance, computer science, and physics. The ability to solve them is critical for understanding how quantities change over time or in reaction to specific conditions.

Rounding Numbers with Precision

Rounding numbers is a useful tool to simplify calculations and present a manageable number of significant figures, especially when dealing with irrational numbers.
When a problem asks you to round to the nearest ten-thousandth, it means you need four decimal places in your answer. In our exercise, after calculating \( x = 2.7183 - 1 = 1.7183 \), the result is already at the needed precision. However, whenever rounding is necessary:

• Look at the number in the position right after your rounding spot. If it's 5 or more, round up. If it's less than 5, round down.

This skill ensures that reports are easy to read and understand. Rounding with precision can prevent errors in computations and is commonly used in fields like engineering, economics, and statistics.