Problem 89 - A Graphical Approach to Precalculus with Limits

Question

The given equations are quadratic in form. Solve each and give exact solutions. $$\left(\log _{2} x\right)^{2}+\log _{2} x=2$$

Step-by-Step Solution

Answer

The exact solutions are \( x = 2 \) and \( x = \frac{1}{4} \).

Step 1: Identifying the Quadratic Form

The equation \( (\log_2 x)^2 + \log_2 x - 2 = 0 \) is in the quadratic form \( ax^2 + bx + c = 0 \). We identify \( a = 1 \), \( b = 1 \), and \( c = -2 \), where \( x = \log_2 x \). This can be treated as a standard quadratic equation by substituting \( y = \log_2 x \).

Step 2: Substituting and Reformulating

Substitute \( y = \log_2 x \) into the equation, so it becomes \( y^2 + y - 2 = 0 \). This is a standard quadratic equation in terms of \( y \).

Step 3: Solving the Quadratic Equation

Apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For the equation \( y^2 + y - 2 = 0 \), substitute \( a = 1 \), \( b = 1 \), and \( c = -2 \) to get \( y = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} \). This gives two solutions: \( y = 1 \) and \( y = -2 \).

Step 4: Back-Substituting to Solve for x

Substitute back \( y = \log_2 x \) to find \( x \). For \( y = 1 \), \( \log_2 x = 1 \) implies \( x = 2^1 = 2 \). For \( y = -2 \), \( \log_2 x = -2 \) implies \( x = 2^{-2} = \frac{1}{4} \).

Step 5: Verification of Solutions

Verify the solutions by substituting back into the original equation. For \( x = 2 \), \( \log_2 2 = 1 \) and satisfies \((1)^2 + 1 = 2\). For \( x = \frac{1}{4} \), \( \log_2 \frac{1}{4} = -2 \) and satisfies \((-2)^2 - 2 = 2\). Both solutions are correct.

Key Concepts

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They help us solve equations where the variable is an exponent. The logarithmic function converts multiplication into addition, turning complex calculations into simpler ones. In our example, the base is 2, which means we are dealing with a binary logarithm:

• The logarithm can be written as \( \log_2 x \). This means the power to which 2 must be raised to get \( x \).
• For instance, \( \log_2 8 = 3 \) because \( 2^3 = 8 \).

Understanding the properties of logs:

• \( \log_b (mn) = \log_b m + \log_b n \)
• \( \log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n \)
• \( \log_b (m^n) = n \cdot \log_b m \)

Using these rules, complicated calculations involving powers become manageable. In this exercise, using \( y = \log_2 x \) transformed a log equation into a quadratic form, making it solvable using familiar algebraic techniques.

Quadratic Formula

The quadratic formula is a universal tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula helps find the values of \( x \) for which the quadratic is zero.The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's a quick breakdown of how to use it:

• Identify the coefficients \( a \), \( b \), and \( c \) from your equation.
• Plug these values into the formula to calculate the solutions.

In our example, the quadratic equation \( y^2 + y - 2 = 0 \) uses \( a = 1 \), \( b = 1 \), and \( c = -2 \), which we substituted into the formula to find \( y \). The results were \( y = 1 \) and \( y = -2 \). These are considered roots or solutions of the quadratic equation. After obtaining these solutions, one can "translate" them back to solve the original logarithmic equation.

Algebraic Solutions

Algebraic solutions involve manipulating equations and expressions to find solutions. This requires a good grasp of algebraic principles such as substitution, factorization, and back-solving.In our given problem:

• We started by substituting \( y = \log_2 x \) to simplify the equation to a recognizable quadratic form.
• This allowed us to apply the quadratic formula directly to solve it.

Once the solutions \( y = 1 \) and \( y = -2 \) were obtained:

• Back-substitute each value into \( y = \log_2 x \).
• Solving gives \( x = 2^1 = 2 \) for \( y = 1 \), and \( x = 2^{-2} = \frac{1}{4} \) for \( y = -2 \).

These steps verified that both values satisfy the original equation, demonstrating how algebra can be applied to not only solve the problem but also confirm the accuracy of the solution. Algebraic methods like these are fundamental in mathematics for solving real-world problems.