Problem 18 - Precalculus

Question

Solve the equation analytically. $$ \ln (x+1)-\ln (x)=3 $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{-1}{1 - e^3}\).

Step 1: Apply Logarithmic Properties

Use the property of logarithms that states \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \), to combine the logarithms on the left-hand side of the equation. This gives: \[ \ln\left(\frac{x+1}{x}\right) = 3 \]

Step 2: Exponentiate Both Sides

To eliminate the natural logarithm, exponentiate both sides of the equation. This changes the equation to: \[ \frac{x+1}{x} = e^3 \] where \(e\) is the base of natural logarithms.

Step 3: Solve for x

Rearrange the equation from Step 2 to solve for \(x\). Multiply both sides by \(x\) to get: \[ x+1 = x \cdot e^3 \] Then, rearrange to isolate \(x\): \[ x + 1 - x \cdot e^3 = 0 \] which simplifies to: \[ x(1 - e^3) = -1 \] Solve for \(x\): \[ x = \frac{-1}{1 - e^3} \]

Key Concepts

Understanding the Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a special type of logarithm with a base \( e \), where \( e \approx 2.71828 \). It is the inverse function of the exponential function \( e^x \). In simpler terms, the natural logarithm answers the question: "To what power should \( e \) be raised to get \( x \)?"

• Inverse Relationship: Remember, if \( y = \ln(x) \), then \( x = e^y \).
• Domain: The function is only defined for positive values of \( x \) (i.e., \( x > 0 \)).
• Range: The output can be any real number \(-\infty < y < \infty\).

Natural logarithms are commonly used because equations involving growth and decay, like in biology and carbon dating, often use \( e \). In our problem,\( \ln(x+1) - \ln(x) = 3 \), it helps simplify complex expressions through its useful properties.

Exploring Exponential Functions

Exponential functions are a crucial concept tightly linked to logarithms. They involve raising a constant base, like \( e \), to a variable exponent. The general form is \( f(x) = e^x \).

• Base: In the exponential function \( e^x \), \( e \) is the base, similar to the base of the natural logarithm.
• Behavior: As \( x \) increases, \( e^x \) grows rapidly since it's exponential. When \( x \) decreases, \( e^x \) approaches zero but never becomes negative.
• Relationship: They counteract \( \ln \), as we see in the solution where exponentiating eliminates \( \ln \).

In the equation from the exercise, after applying logarithmic properties, we use the notion that exponentiating both sides (\( e^{\ln{a}} = a \)) helps isolate the unknown \( x \). This allows us to focus purely on manipulating the expression algebraically.

Mastering Algebraic Manipulation

Algebraic manipulation involves reworking an equation to make it simpler or to solve for a specific variable.

• Start with Simplification: Use logarithmic properties to merge the terms on the left-hand side of our original equation.
• Isolate Variables: When trying to solve for \( x \), move all \( x\)-terms to one side of the equation. This often requires operations like addition, subtraction, and division.
• Check Solution: Substitute your solution back into the original equation if possible to verify correctness.

In our exercise, after exponentiating, we rearrange the terms to solve for \( x \): Multiply through by \( x \), then factor out \( x \) to make it more apparent. The final rearrangement and division give us \( x = \frac{-1}{1-e^3} \). This showcases our prowess with algebraic techniques, which is fundamental for solving many mathematical problems.