Problem 81 - Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ 2 \log _{2} x=3+\log _{2}(x-2) $$

Step-by-Step Solution

Answer

The solution to the equation is exactly \( x = 4 \).

Step 1: Simplify the equation

The original equation is \( 2 \log_2 x = 3 + \log_2 (x-2) \). We can use the properties of logarithms to simplify it by combining the logs on the left-hand side. Use the power rule for logarithms: \( c \log_b a = \log_b (a^c) \), which gives \( \log_2 (x^2) \). Thus, the equation becomes \( \log_2 (x^2) = 3 + \log_2 (x-2) \).

Step 2: Use the property of equality in logarithms

Since we have \( \log_2 (x^2) = 3 + \log_2 (x-2) \), we need to isolate the logarithms on one side to use the property that \( \log_b m = \log_b n \Rightarrow m = n \). Subtract \( \log_2 (x-2) \) from both sides: \( \log_2 (x^2) - \log_2 (x-2) = 3 \).

Step 3: Apply the quotient rule for logarithms

Use the property \( \log_b a - \log_b c = \log_b \left( \frac{a}{c} \right) \) on the equation \( \log_2 (x^2) - \log_2 (x-2) = 3 \). This transforms the equation to \( \log_2 \left( \frac{x^2}{x-2} \right) = 3 \).

Step 4: Convert the logarithmic equation to an exponential equation

The equation \( \log_2 \left( \frac{x^2}{x-2} \right) = 3 \) can be rewritten in exponential form, \( \frac{x^2}{x-2} = 2^3 \); thus, \( \frac{x^2}{x-2} = 8 \).

Step 5: Solve the rational equation

Multiply both sides by \( x-2 \) to eliminate the fraction: \( x^2 = 8(x-2) \). Simplify to get \( x^2 = 8x - 16 \). Rearrange this to get the quadratic equation: \( x^2 - 8x + 16 = 0 \).

Step 6: Solve the quadratic equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = -8, c = 16 \). Calculate the discriminant: \( b^2 - 4ac = (-8)^2 - 4 \cdot 1 \cdot 16 = 64 - 64 = 0 \). Since the discriminant is zero, there is one real solution: \( x = \frac{-(-8) \pm \sqrt{0}}{2 \cdot 1} = \frac{8}{2} = 4 \).

Step 7: Check for extraneous solutions

Logarithms are only defined for positive numbers. Since \( x = 4 \) and the original equation involved \( \log_2 (x-2) \) which becomes \( \log_2 2 \) when \( x = 4 \), there are no issues with this value. Thus, \( x = 4 \) is a valid solution.

Key Concepts

Logarithmic Equations

Logarithmic equations have variables within the logarithm and often appear in the form \( \log_b(x) = n \), where \( b \) is the base and \( n \) indicates the power to which the base \( b \) must be raised to result in \( x \). Solving these equations entails finding that unknown variable.

To solve a logarithmic equation like \( 2 \log_2 x = 3 + \log_2 (x-2) \), you can utilize properties of logarithms to simplify and solve for \( x \). The power rule allows you to rewrite \( 2 \log_2 x \) as \( \log_2(x^2) \). This equation, once simplified using other properties, reveals an expression that can be converted to its equivalent exponential form, which is often easier to solve.

When dealing with logarithmic equations, always consider if solutions are extraneous or do not fit the constraints of logarithmic definitions. Since logarithms require positive arguments, verify that your solutions do not result in negative or zero inside the log function.

Quadratic Equations

A quadratic equation is the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. Solving these equations involves factoring, completing the square, or using the quadratic formula.

In our problem, simplifying the logarithmic equation led to the quadratic equation \( x^2 - 8x + 16 = 0 \). This quadratic can be solved using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• For this equation, \( a = 1 \), \( b = -8 \), \( c = 16 \)
• The discriminant, \( b^2 - 4ac \), is zero, indicating one real repeated solution

The solution, \( x = 4 \), is consistent and requires verification through substitution into the original equation. Quadratics often yield multiple solutions, but the nature of the discriminant here warrants a single valid value for \( x \).

Properties of Logarithms

Understanding the properties of logarithms makes solving logarithmic equations more approachable. Several key properties include:

• Product Rule: \( \log_b(mn) = \log_b m + \log_b n \)
• Quotient Rule: \( \log_b\left(\frac{m}{n}\right) = \log_b m - \log_b n \)
• Power Rule: \( \log_b(m^n) = n \cdot \log_b m \)

The given problem \( 2 \log_2 x = 3 + \log_2(x-2) \) utilizes these properties. By applying the power rule, \( 2 \log_2 x \) is transformed into \( \log_2(x^2) \). The equation can then be simplified by using the quotient rule, turning \( \log_2(x^2) - \log_2(x-2) = 3 \) into \( \log_2\left(\frac{x^2}{x-2}\right) = 3 \). This simplification facilitates converting the log equation to its exponential form, enhancing ease of solution. These properties are foundational tools for manipulating and solving complex logarithmic equations effectively.