Problem 58 - A Graphical Approach to Precalculus with Limits

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$\log _{2}(2 x)+\log _{2}(x+2)=\log _{2} 16$$

Step-by-Step Solution

Answer

x = 2

Step 1: Apply Logarithm Property

The given equation is \( \log_{2}(2x) + \log_{2}(x+2) = \log_{2}16 \). Use the property that \( \log_{b}(A) + \log_{b}(B) = \log_{b}(A \cdot B) \) to combine the left side into one logarithm: \[ \log_{2}((2x)(x+2)) = \log_{2}16 \]

Step 2: Simplify the Equation

Since the logarithms are equal, their arguments must be equal. Set the arguments equal to each other: \[ (2x)(x+2) = 16 \].

Step 3: Expand and Solve the Quadratic Equation

Expand the left-hand side: \( 2x(x+2) = 2x^2 + 4x \). The equation becomes \[ 2x^2 + 4x = 16 \]. Subtract 16 from both sides to form a quadratic equation: \[ 2x^2 + 4x - 16 = 0 \].

Step 4: Factor the Quadratic Equation

First, divide the entire equation by 2 to simplify: \( x^2 + 2x - 8 = 0 \). Now, factor it: \[ (x+4)(x-2) = 0 \].

Step 5: Solve for x

Set each factor equal to zero and solve for \( x \): \( x+4 = 0 \) or \( x-2 = 0 \). This gives \( x = -4 \) or \( x = 2 \).

Step 6: Check for Extraneous Solutions

Since logarithms are only defined for positive numbers, check the solutions in the context of the original logarithmic expressions. The solution \( x = -4 \) results in a negative argument inside the logarithm, making it invalid. Thus, only \( x = 2 \) is valid.

Key Concepts

Quadratic Equations

Quadratic equations form a crucial part of algebra, dealing with polynomial expressions where the highest power of the variable is two. In this exercise, after applying logarithm properties, the equation is transformed into a quadratic equation:\[ 2x^2 + 4x - 16 = 0 \]To simplify solving, divide by 2:\[ x^2 + 2x - 8 = 0 \]This equation is now in its standard form, which is \( ax^2 + bx + c = 0 \). Solving these involves factoring, completing the square, or using the quadratic formula.- Factoring: Express the quadratic as a product of two linear expressions. For \( x^2 + 2x - 8 \), it factors to: \[ (x + 4)(x - 2) = 0 \] Solving each factor, we get two possible solutions: \( x = -4 \) and \( x = 2 \).Understanding how to handle quadratic equations is essential in solving various algebraic problems.

Logarithm Properties

Logarithmic equations often require the application of logarithm properties to simplify and solve them. A key property used in this exercise is:\[ \log_b(A) + \log_b(B) = \log_b(A \times B) \]This allows the combination of multiple logarithms into a single expression, simplifying the equation.Here, we combined the logarithms on the left:\[ \log_{2}((2x)(x+2)) = \log_{2}16 \]The equality of logarithms implies their arguments must be equal, leading to:\[ (2x)(x+2) = 16 \]These properties are extremely useful for solving and understanding the behavior of logarithmic equations. Knowing when and how to apply them helps transform complex expressions into manageable forms.

Extraneous Solutions

Extraneous solutions are results that emerge from solving an equation but do not satisfy the original equation after verification. In scenarios involving logarithms, the natural restrictions on the values of inputs (arguments of the log) often lead to such solutions.Logarithms are defined only for positive arguments. Hence, when solving \( \log_2(2x) + \log_2(x+2) = \log_2 16 \), we must ensure that the solutions - Make \( 2x > 0 \)- Make \( x + 2 > 0 \)After finding the solutions \( x = -4 \) and \( x = 2 \), we check them against these conditions:- \( x = -4 \) leads to negative arguments, as \( 2(-4) = -8 \) and \(-4 + 2 = -2 \). It's invalid.- \( x = 2 \) satisfies both conditions, \( 2 \times 2 = 4 \) and \( 2 + 2 = 4 \), making it valid.Remembering to verify solutions in the context of the initial problem ensures we accept results that genuinely solve the original equations.