Problem 87 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) 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. $$ \log x+\log (x+3)=\log 10 $$

Step-by-Step Solution

Answer

The solution to the given logarithmic equation \(\log x+ \log (x+3)= \log 10\) is \(x = 2\)

Step 1: Simplify the Equation

By using the properties of logarithms, the first step is to combine the logarithms on the left side of the equation. In detail, using the product rule which states \(\log b + \log c = \log (b \cdot c)\), the sum of the two logarithms can be written as a single logarithm: \(\log x+\log (x+3) = \log (x \cdot (x+3)) = \log (x^2+3x)\) . So, the equation becomes \(\log (x^2 + 3x) = \log 10\).

Step 2: Equivalency of Logarithms

Since the two expressions are both base-10 logarithms, by the equivalency of logarithms attributing to the same base, the argument of the logarithms can be set equal to each other yielding \(x^2 + 3x = 10\) .

Step 3: Solve the Quadratic Equation

To solve the equation \(x^2 + 3x = 10\), we rearrange it to make the equation a standard form of a quadratic equation by subtracting 10 from both sides. Therefore, the formula changes to \(x^2 + 3x - 10 = 0\)

Step 4: Factorization of the Quadratic equation

This resulting quadratic equation can be factored: \(x^2 + 3x - 10 = 0\) can be factored to \((x+5)(x-2)= 0\)

Step 5: Find root of the equation

Setting the factored equation equal to zero, we have two solutions for \(x\), they are \(x = -5\) and \(x = 2\).

Step 6: Check the Domain

In logarithms, the domain is \(x > 0\), which means x should be a positive number. Therefore, although \(x=-5\) is a solution to the equation, it needs to be rejected because of the domain constraint of the original logarithmic expression.

Key Concepts

Properties of Logarithms

Logarithms follow specific rules that help simplify expressions. These properties are incredibly useful when solving logarithmic equations. Let's look at the most commonly used properties:

• Product Rule: \( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \). This is used to combine the sum of two logs into one single log with a product inside.
• Quotient Rule: \( \log_b(m) - \log_b(n) = \log_b \left( \frac{m}{n} \right) \). This allows you to subtract logarithms by creating a single log with a division inside.
• Power Rule: \( \log_b(m^n) = n \cdot \log_b(m) \). This is a way to bring the exponent in front of the log as a multiplier.

In solving the original problem, the product rule was particularly important. It allowed us to combine \( \log x + \log (x+3) \) into \( \log(x \cdot (x+3)) \), simplifying the expression to \( \log(x^2 + 3x) \). This makes the equation easier to work with.

Quadratic Equations

A quadratic equation is any equation that can be rearranged in this general form: \( ax^2 + bx + c = 0 \). It looks like a parabola when graphed and often presents itself in problems involving squared terms.
The most common methods to solve quadratic equations include:

• Factoring: This is possible when the quadratic can be expressed as the product of two binomials.
• Quadratic Formula: Given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), and used when factoring is difficult or impossible.
• Completing the Square: Useful in converting the quadratic into a perfect square trinomial.

In the provided solution, factorizable quadratic equations allowed us to rewrite \( x^2 + 3x - 10 = 0 \) as \( (x+5)(x-2)= 0 \). This revealed potential solutions for \( x \), which are -5 and 2. However, always remember to verify any restrictions posed by the original equation's domain.

Domain of Logarithmic Functions

Logarithmic functions have specific domain restrictions which are crucial to solving logarithmic equations correctly. Essentially, the argument inside the log must always be positive:

• The expression inside the logarithm, i.e., the argument, should not be zero or negative.
• If \( \log(x) \) is given, then \( x > 0 \).

In solving the problem, we discovered solutions \( x = -5 \) and \( x = 2 \). However, because logarithms forbid zero or negative inputs, only \( x = 2 \) is valid.
It's important to always check if your solution fits the domain. This ensures the answers satisfy not just the numerical solution but also the logical preconditions set by the logarithms themselves. This is why \( x = -5 \) is dismissed in our exercise solution.