Problem 42 - Algebra and Trigonometry

Question

Solve the logarithmic equation for \(x\) $$ \log _{2}\left(x^{2}-x-2\right)=2 $$

Step-by-Step Solution

Answer

The solutions for \( x \) are 3 and -2.

Step 1: Understand the Equation

The given logarithmic equation is \( \log_{2}(x^2 - x - 2) = 2 \). This means we are looking for a value of \( x \) that makes \( x^2 - x - 2 \) equal to \( 2^2 \) since the base of the logarithm is 2.

Step 2: Remove Logarithm by Exponentiation

To eliminate the logarithm, we exponentiate both sides of the equation using the base 2. The equation becomes: \( x^2 - x - 2 = 2^2 \). This simplifies to \( x^2 - x - 2 = 4 \).

Step 3: Set Up a Quadratic Equation

Rearrange the equation to form a quadratic equation: \( x^2 - x - 2 - 4 = 0 \). Simplify this to get \( x^2 - x - 6 = 0 \).

Step 4: Factor the Quadratic Equation

To solve \( x^2 - x - 6 = 0 \), factor the quadratic equation: \( (x - 3)(x + 2) = 0 \).

Step 5: Solve for x

Set each factor equal to zero to find the values of \( x \). If \( x - 3 = 0 \), then \( x = 3 \). Similarly, if \( x + 2 = 0 \), then \( x = -2 \).

Step 6: Check the Solutions in Original Equation

Substitute \( x = 3 \) into the original expression: \( x^2 - x - 2 = 3^2 - 3 - 2 = 4 \). The left side equals 4, which matches \( 2^2 \), so \( x = 3 \) is a valid solution. Substitute \( x = -2 \): \( (-2)^2 - (-2) - 2 = 4 \), which also equals 4, so \( x = -2 \) is a valid solution.

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions where the highest degree of the variable is a square. In simpler terms, it looks something like this: \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero. These equations often have two solutions, which can be found using various methods, including factoring, completing the square, or using the quadratic formula. Quadratic equations are essential in finding solutions related to parabolic paths, such as projectile motion or optimizing certain problems in economics.

Factoring Polynomials

Factoring polynomials involves rewriting a polynomial as a product of its factors. When we factor quadratic equations, like \( x^2 - x - 6 = 0 \), we look for two numbers that multiply to the constant term (in this case, -6) and add up to the linear coefficient (-1 here).

• These numbers are 3 and -2 since \( 3 \times (-2) = -6 \) and \( 3 + (-2) = 1 \).
• The equation can be rewritten as \((x - 3)(x + 2) = 0 \).

Once we have factored the polynomial, solving the equation becomes straightforward by setting each factor equal to zero, simplifying the process of finding \( x \). Reminder: Always check your solutions in the context of the original problem to ensure they make sense numerically and logically.

Exponentiation

Exponentiation is the process of raising a number, known as the base, to a particular power. If we say \( 2^2 \), this means 2 is the base, and we multiply it by itself to get 4. Exponentiation is vital in solving logarithmic equations because the process involves reversing a logarithm. When a logarithmic equation is first presented, such as \( \log_{2}(x^2 - x - 2) = 2 \), exponentiation helps eliminate the logarithm. By applying exponentiation, we retrieve the original expression: \( x^2 - x - 2 \) becomes 4.

Base 2 Logarithms

Base 2 logarithms, also known as binary logarithms, involve finding what power the number 2 must be raised to produce a given value. In the equation \( \log_{2}(x^2 - x - 2) = 2 \), we asked, "To what power must 2 be raised to yield what's inside the log?" The answer is the number 4, because \( 2^2 = 4 \). Base 2 logarithms are common in computer science and information theory due to binary computation. Understanding and manipulating logarithms are crucial in these fields as they simplify the calculation of complex powers and roots. They also often appear in algorithms that involve data complexity and storage requirements. With mastery of logarithms, especially those with base 2, solving equations efficiently becomes much more manageable.