Problem 80 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ \log _{5}(7+x)+\log _{5}(8-x)-\log _{5} 2=2 $$

Step-by-Step Solution

Answer

Exact: \( x = \frac{1 + \sqrt{177}}{2} \); Approximate: \( x \approx 7.1521 \).

Step 1: Combine Logarithms on the Left

Use the properties of logarithms to combine the terms. Remember that \( \log_a b + \log_a c = \log_a (bc) \) and \( \log_a b - \log_a c = \log_a \left(\frac{b}{c}\right) \).\[\log_5((7+x)(8-x)) - \log_5 2 = \log_5 \left( \frac{(7+x)(8-x)}{2} \right) = 2\]

Step 2: Eliminate the Logarithm

Since the logarithms are now simplified, you can solve for \((7+x)(8-x)\) by making both sides of the equation not logarithmic. Use the fact that if \( \log_a b = c \), then \( b = a^c \).\[\frac{(7+x)(8-x)}{2} = 5^2\]Simplify:\[(7+x)(8-x) = 50 \times 2 = 100\]

Step 3: Expand and Solve the Quadratic Equation

First expand the left side and simplify:\[(7+x)(8-x) = 7 \times 8 - 7x + 8x - x^2 = 56 + x - x^2\]Set the equation to zero:\[-x^2 + x + 56 = 100\]Rearrange terms:\[-x^2 + x + 56 - 100 = 0\]Simplify:\[-x^2 + x - 44 = 0\]Multiply through by -1 for simplicity:\[ x^2 - x + 44 = 0 \]

Step 4: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -1, c = -44 \).\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-44)}}{2 \cdot 1}\]\[x = \frac{1 \pm \sqrt{1 + 176}}{2}\]\[x = \frac{1 \pm \sqrt{177}}{2}\]Calculate:\[x_1 = \frac{1 + \sqrt{177}}{2}, \quad x_2 = \frac{1 - \sqrt{177}}{2}\]

Step 5: Evaluate the Solutions

Check for validity. Notice that only \(x_1 = \frac{1 + \sqrt{177}}{2}\) makes both \(7 + x > 0\) and \(8 - x > 0\) valid. \(x_2 = \frac{1 - \sqrt{177}}{2}\) results in a negative value, making \(7+x > 0\) invalid. Hence, only \( x_1 \) is valid. Approximately, \( \sqrt{177} \approx 13.3041 \), so:\[x_1 \approx \frac{1 + 13.3041}{2} \approx 7.15205\]

Step 6: Provide Exact and Approximate Solutions

The exact solution to the quadratic equation that satisfies the original logarithmic equation is \( x = \frac{1 + \sqrt{177}}{2} \). The approximate solution, to four decimal places, is \( x \approx 7.1521 \).

Key Concepts

Quadratic Equations

Understanding quadratic equations can greatly benefit your mathematical journey. These equations often appear in various forms and are typically written as:

• \( ax^2 + bx + c = 0 \)

Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \).
Quadratic equations arise naturally in many problems, especially when dealing with products or areas, such as when we expanded the logarithmic terms in this exercise. When solving these equations, a useful tool is the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

The expression under the square root, \( b^2 - 4ac \), is called the discriminant. It helps us determine the nature of the roots:

• If it's positive, there are two distinct real roots.
• If zero, there's exactly one real root (a repeated root).
• If negative, the roots are complex and not real numbers.

In our solution, the discriminant was positive, resulting in two real roots, but only one valid solution for the problem context.

Properties of Logarithms

Logarithms are powerful mathematical tools, especially when working with exponential relationships. The properties of logarithms can simplify complex expressions, as seen in solving the given logarithmic equation. Here are some essential properties:

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

These properties enable us to transform multiplicative relationships into additive ones, simplifying the process of solving. In the problem, we used these properties to combine multiple logarithms into a single expression, making it easier to equate and solve. Remember, understanding these properties is key to mastering logarithmic equations.

Approximation Methods

After obtaining the exact solutions, it is common to approximate results for easier interpretation or practical use. Here, we found the exact solution using the quadratic formula and then sought an approximate decimal value.
This involves estimating the square roots and performing basic arithmetic to achieve a solution accurate to four decimal places. Calculators or computational tools are usually employed for such tasks since they ensure precision.
Approximations are especially valuable in real-world scenarios where measurements and calculations can't always be precise. By finding both exact and approximate solutions, we cater to theoretical exploration and practical application. In this problem, while the exact solution was \( x = \frac{1 + \sqrt{177}}{2} \), the approximate value, calculated as 7.1521, offers a more tangible measure to work with in everyday situations.