Problem 88 - Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \frac{1}{2} \log (4 x+5)=\log x $$

Step-by-Step Solution

Answer

The exact solution to the equation is \( x = 5 \).

Step 1: Eliminate the Fraction from the Logarithm

We start with the equation \( \frac{1}{2} \log(4x + 5) = \log x \). To remove the fraction \( \frac{1}{2} \), we multiply both sides of the equation by 2, resulting in \[ \log(4x + 5) = 2 \log x. \]

Step 2: Apply the Power Rule of Logarithms

The equation \( \log(4x + 5) = 2 \log x \) can be rewritten using the power rule of logarithms, \( \log(x^a) = a \log x \). Applying this rule, we have \[ \log(4x + 5) = \log(x^2). \]

Step 3: Set the Arguments Equal

Since \( \log(4x + 5) = \log(x^2) \), we can set the arguments equal: \( 4x + 5 = x^2 \). This gives us the quadratic equation \[ x^2 - 4x - 5 = 0. \]

Step 4: Solve the Quadratic Equation

We solve \( x^2 - 4x - 5 = 0 \) using the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -4 \), and \( c = -5 \). Substitute the values: \( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} \). Simplifying further, \[ x = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2}. \]

Step 5: Simplify the Solutions

The expression \( \sqrt{36} = 6 \), so the solutions become \( x = \frac{4 + 6}{2} \) and \( x = \frac{4 - 6}{2} \). Solving these gives \( x = 5 \) and \( x = -1 \).

Step 6: Verify Solutions

Since \( x = -1 \) is not valid in a logarithmic expression (where \( x \) must be positive), we discard it. Therefore, the only valid solution is \( x = 5 \).

Step 7: Approximation

The solution \( x = 5 \) is already an exact value. If approximation is required, \( x = 5 \) remains the same to four decimal places.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). With this formula, you can solve any quadratic equation. You'll often encounter these in various math problems. The formula is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here's how it works:

• \( a \), \( b \), and \( c \) are coefficients from the quadratic equation.
• The symbol \( \pm \) indicates that there will generally be two solutions for \( x \).
• The term \( b^2 - 4ac \) is called the discriminant, which determines the nature of the roots.

- A positive discriminant (\( > 0 \)) indicates two real and distinct solutions.- A zero discriminant (\( = 0 \)) indicates one real solution (a repeated root).- A negative discriminant (\( < 0 \)) means there are two complex solutions.Using the quadratic formula, we solved \( x^2 - 4x - 5 = 0 \) in the exercise. This resulted in values of \( x = 5 \) and \( x = -1 \). However, because logarithmic functions require positive values, we only considered \( x = 5 \).

Logarithmic Properties

Logarithmic properties are essential tools in simplifying complex logarithmic equations. These properties include the product rule, quotient rule, and power rule:

• 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: \( \log_b(M^a) = a \log_b M \)

In the original problem, we used the power rule to transition from \( 2 \log x \) to \( \log(x^2) \). Understanding these rules allows you to manipulate and solve logarithmic equations efficiently. By applying the power and equality properties, we set \( \log(4x+5) \) equal to \( \log(x^2) \), making it possible to transition from a logarithmic to a quadratic equation. This step was crucial for reaching the solution of the exercise.

Exact Solutions

Exact solutions refer to the precise answer obtained without any approximations. These are particularly important in mathematical problems where accuracy is imperative. In our exercise with the logarithmic equation, we derived an exact solution of \( x = 5 \). Why are exact solutions vital? Because:

• They provide a clear and definitive answer.
• In many mathematical contexts, an exact answer is necessary to ensure correctness.
• Exact solutions give insight into mathematical relationships without rounding errors.

The step-by-step approach ensured that we derived \( x = 5 \) correctly without needing approximations, which highlights the value of exactness in problem-solving.

Approximations in Mathematics

Approximations are often used when exact solutions are difficult or not possible to obtain. They provide a way to work with large numbers or complex equations. In practical scenarios, approximations help when you need near solutions quickly. However, in our case:

• The solution \( x = 5 \) was already exact.
• If approximations were required, \( x = 5.0000 \) to four decimal places would suffice.

Approximations come in handy especially in real-world applications like engineering and physics where measurements can rarely be perfect. It’s critical to note, though, that precision is key. Over-reliance on approximations without a grasp of exact solutions might lead to errors in results. Thus, understanding when and how to approximate, particularly when your starting solution is already exact, like in our logarithmic problem, is essential.