Problem 50 - Algebra and Trigonometry

Question

Solve the logarithmic equation for \(x\) $$ \ln (x-1)+\ln (x+2)=1 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{-1 + \sqrt{9 + 4e}}{2} \).

Step 1: Use the logarithm property

Start with the given equation: \[ \ln(x-1) + \ln(x+2) = 1 \]Use the logarithm property \( \ln a + \ln b = \ln(ab) \) to combine the logs:\[ \ln\big((x-1)(x+2)\big) = 1 \]

Step 2: Exponentiate both sides

Exponentiate both sides to eliminate the natural logarithm:\[ (x-1)(x+2) = e^1 \]Simplify the right side:\[ (x-1)(x+2) = e \]

Step 3: Expand the equation

Expand the left side of the equation:\[ x^2 + 2x - x - 2 = e \]Combine like terms:\[ x^2 + x - 2 = e \]

Step 4: Set up the quadratic equation

Set up the equation as a standard quadratic equation:\[ x^2 + x - (2 + e) = 0 \]

Step 5: Solve using the quadratic formula

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = 1\), \(b = 1\), and \(c = -(2+e)\):\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -(2+e)}}{2 \cdot 1} \]Simplify the expression under the square root:\[ x = \frac{-1 \pm \sqrt{1 + 8 + 4e}}{2} \]\[ x = \frac{-1 \pm \sqrt{9 + 4e}}{2} \]

Step 6: Evaluate the solutions

Solve for both potential solutions of \(x\):1. \(x = \frac{-1 + \sqrt{9 + 4e}}{2} \)2. \(x = \frac{-1 - \sqrt{9 + 4e}}{2} \)Evaluate which values are valid within the context (since \(x - 1 > 0\) and \(x + 2 > 0\), \(x > 1\) and \(x > -2\)): Therefore, \(x = \frac{-1 + \sqrt{9 + 4e}}{2} \) is the valid solution as it satisfies the conditions. The second solution is discarded as it results in a negative value.

Key Concepts

Natural Logarithm

The natural logarithm, often written as \( \ln(x) \), is logarithm taken to the base of the mathematical constant \( e \), which is approximately 2.718. The function \( \ln(x) \) is the inverse of the exponential function \( e^x \).
This means that if you take \( e \) raised to the power of \( \ln(x) \), you will get back \( x \).
Understanding the natural logarithm is crucial because it appears frequently in problems involving exponential growth and decay, as well as in calculus and differential equations.
Here's a quick rundown of some foundational ideas related to \( \ln(x) \):

• It is only defined for positive numbers.
• \( \ln(1) = 0 \), because \( e^0 = 1 \).
• \( \ln\) transforms multiplication into addition: \( \ln(xy) = \ln(x) + \ln(y) \).
• For constants \( a \), \( \ln(x^a) = a \ln(x) \).

Properties of Logarithms

Logarithmic properties are tools that simplify calculations and helps solve logarithmic equations efficiently. In the given exercise, the property \( \ln(a) + \ln(b) = \ln(ab) \) plays a crucial role.
Here is a breakdown of the most commonly used properties:

• **Product Rule**: \( \ln(ab) = \ln(a) + \ln(b) \). It turns multiplication inside a logarithm into addition outside of it.
• **Quotient Rule**: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \). It converts division inside a logarithm to subtraction outside.
• **Power Rule**: \( \ln(a^b) = b \ln(a) \). It brings down the exponent as a multiplier.

Using these properties lets you combine, separate, or simplify complex logarithmic expressions, which is extremely useful when solving equations or analyzing mathematical models.

Quadratic Formula

The quadratic formula solves quadratic equations of the form \( ax^2 + bx + c = 0\). In this task, the equation was brought down to this form after applying logarithmic properties and exponentiation.
The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here's a breakdown of key components:

• **\(a, b, c\)**: These are the coefficients of the quadratic equation, with \(a\) not equal to zero.
• **Discriminant \(b^2 - 4ac\)**: This part of the formula, inside the square root, determines the nature of the roots. If it is positive, you get two distinct real roots; if zero, exactly one real root; and if negative, no real roots but rather complex ones.

The advantage of the quadratic formula is its universal applicability to solve any quadratic equation, making it an essential tool for students.

Exponentiation

Exponentiation is a fundamental operation in mathematics, involving repeated multiplication of the same number. In the context of the given exercise, exponentiation is used to eliminate the natural logarithm from an equation.
The process can be identified as follows:

• Starting from \( \ln(x) = y \), applying exponentiation creates \( e^y = x \).
• In equations: If \( \ln(f(x)) = c \), then by exponentiating it transforms into \( f(x) = e^c \).

Exponentiation not only undoes the natural log but also plays a vital role in growth patterns and scientific calculations, such as calculating compound interest or modeling population growth. Understanding how to properly apply or reverse exponentiation delivers powerful insight into handling exponential functions efficiently.