Problem 67 - Introductory and Intermediate Algebra for College Students 4th

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. $$\log _{3}(x-5)+\log _{3}(x+3)=2$$

Step-by-Step Solution

Answer

Therefore, the exact solution to the given logarithmic equation is \(x=6\)

Step 1: Combine Logarithms

Using the principle of addition in logarithms, the given equation can be rewritten by combining the two logarithms with the same base into one:\( \log_{3}((x-5)(x+3)) = 2 \)

Step 2: Convert Logarithm to Exponential Form

In the next step, convert the logarithmic equation into an exponential equation:\( 3^2 = (x-5)(x+3) \)

Step 3: Simplify the Equation

Simplify the equation by expanding the terms on the right and subtracting 9 from each side to set it to 0:\( x^2 -2x - 24 = 0 \)

Step 4: Solve the Quadratic Equation

Solve this quadratic equation using the quadratic formula \(x=(-b±\sqrt{b^2-4ac})/2a\). In this equation, a=1, b=-2, c=-24. Solving yields:\(x=6,-4\)

Step 5: Verify the Domain

However, the domain of the original logarithmic expression and the resulting quadratic should be the same. Both \(x-5\) and \(x+3\) must be greater than 0. Thus, x>5 and x>-3, so the only valid solution which is in the domain of the original problem is \(x=6\).

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is essential to solving logarithmic equations. The logarithmic functions operate with specific rules that can be used to manipulate and simplify expressions. One key property is that the logarithm of a product is the sum of the logarithms when they have the same base. This means that
\(\log_b(x) + \log_b(y) = \log_b(xy)\).
This property is useful because it allows us to combine logarithmic terms into a single term, simplifying the equation and making it easier to solve. Another important property is the power rule, which states that
\(\log_b(x^a) = a \cdot \log_b(x)\),
allowing exponents to be brought in front of the log as multipliers. By applying these properties, we can reshape logarithmic equations and prepare them for conversion to their exponential forms for easier solving, as seen in the textbook exercise provided.

Exponential Form Conversion

To solve logarithmic equations, it's often necessary to convert them into their exponential form. This is because it's easier to deal with exponents than with logs when it comes to finding the unknown values. The conversion is based on the fundamental definition of a logarithm:
if \(\log_b(a) = c\), then \(b^c = a\).
Applying this definition allows you to switch from a logarithmic equation to an exponential one, which often leads to a more straightforward problem. For example, the equation
\(\log_3(x-5) + \log_3(x+3) = 2\)
was converted to \(3^2 = (x-5)(x+3)\) by utilizing this conversion. This transformation is a crucial step in finding the solution, as it sets the stage for solving a quadratic equation, which is a well-established process.

Quadratic Equations

Quadratic equations, which take the general form \(ax^2 + bx + c = 0\), can be solved using various methods, such as factoring, completing the square, or employing the quadratic formula.
For many students, the quadratic formula, given as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), is a reliable tool because it can handle any quadratic equation, even those that are not easily factored. In our original exercise, after converting the problem into exponential form and simplifying, we obtained a quadratic equation that we then solved using the quadratic formula. It's crucial to note that when solving for \(x\), it's possible to get two solution candidates, but it's important to verify which, if any, are within the domain of the original logarithmic equation.
In our exercise, we found two solutions, \(x = 6\) and \(x = -4\), but only \(x = 6\) was within the required domain, illustrating the need to validate solutions against the domain constraints to arrive at the correct final answer.