Problem 88 - Algebra and Trigonometry

Question

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log (x+3)+\log (x-2)=\log 14 $$

Step-by-Step Solution

Answer

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

Step 1: Combine Logarithms

Use the property of logarithms \(\log a + \log b = \log ab\) to combine the two logarithms on the left-hand side of the equation. By using this property, \(\log (x+3)+\log (x-2)\) becomes \(\log((x+3)(x-2))\). So, the equation \(\log (x+3)+\log (x-2)=\log 14\) becomes \(\log((x+3)(x-2))=\log 14\).

Step 2: Remove Logarithms

Since the bases of the logarithms on both sides of the equation are the same, we can set the arguments equal to each other. Hence, \((x + 3)(x - 2) = 14\).

Step 3: Solve for X

Expand the left-hand side and subtract 14 from both sides. This results in a quadratic equation: \(x^2 + x - 20 = 0\). This can be factored into \((x - 4)(x + 5)=0\). So, the solutions for \(x\) are \(x = 4\) and \(x = -5\).

Step 4: Check for domain restrictions

However, we must remember that the original logarithmic expressions are not defined for all real numbers. We must reject the solution \(x = -5\), as it would yield negative arguments for both \(\log (x+3)\) and \(\log (x-2)\) in the original equation. The solution \(x = 4\) is valid as it yields positive arguments for both original logarithms.

Key Concepts

Domain of Logarithms

Understanding the domain of logarithms is crucial for solving logarithmic equations. Logarithmic functions, such as \(\log(x)\), are only defined for positive numbers. This means that the argument inside the logarithm, \(x\), must always be greater than zero. If \(x\) is negative or zero, the logarithm becomes undefined. In the context of our exercise, when given \(\log(x+3)+\log(x-2)=\log 14\), the expressions \(x+3\) and \(x-2\) must both be positive for the logarithms to be valid.

• For \(\log(x+3)\): \(x+3>0\) which implies \(x>-3\).
• For \(\log(x-2)\): \(x-2>0\) which implies \(x>2\).

Thus, the combined domain for this equation mandates that \(x>2\). Solving the quadratic equation may yield multiple solutions, but it is crucial to check each solution within this domain to ensure the argument of each logarithmic expression remains positive.

Properties of Logarithms

The properties of logarithms are essential tools that simplify solving equations involving logarithms. Among these properties, the product property is particularly useful for combining and solving logarithmic expressions. It states that \(\log a + \log b = \log(ab)\). This allows us to consolidate terms and tackle equations more efficiently.For example, in the given exercise:- \(\log(x+3) + \log(x-2) = \log 14\)we can use the product property to combine the left-hand side of the equation:- \(\log((x+3)(x-2)) = \log 14\)Once logs with the same base are consolidated, we can safely remove the logarithms and set the arguments equal:- \((x+3)(x-2) = 14\)This step simplifies the equation into a form that can be solved algebraically without logarithms.

Quadratic Equations

Quadratic equations are a polynomial expression of the form \(ax^2 + bx + c = 0\). Solving quadratic equations can often involve factoring, which simplifies complex expressions into simpler, multiplicative forms involving \(x\).After using properties of logarithms in the exercise, we land at the equation:- \((x+3)(x-2) = 14\)Expanding and rearranging gives us the standard quadratic form:- \(x^2 + x - 20 = 0\)This can be factored into:- \((x - 4)(x + 5) = 0\)Setting each factor equal to zero provides the solutions \(x = 4\) and \(x = -5\). However, solution \(x = -5\) is outside our domain requirement for the logarithms and must be rejected. The valid solution is \(x = 4\), which falls within the allowable domain, ensuring positive arguments for the original logarithmic terms.