Problem 70 - 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. $$ \log _{6}(x+3)+\log _{6}(x+4)=1 $$

Step-by-Step Solution

Answer

The given logarithmic equation has no solution. Even though \(x = -1\) and \(x = -6\) make the equation true, both results are not in the domain of the original logarithmic expression, as logarithms are only defined for positive numbers.

Step 1: Application of Logarithm Properties

Use the property of logarithms that says loga(x) + loga(y) = loga(xy) to combine the two log terms on the left hand side of the equation: \\(\log _{6}(x+3)+\log _{6}(x+4) =\log_{6}[(x+3)(x+4)] = 1\)\\

Step 2: Convert Logarithmic Equation to Exponential Equation

The logarithmic equation can be converted to an exponential equation by rewriting the equation as \\b = a^(\log_a(b)). Therefore, \\(x+3)(x+4) = 6^1 = 6\)\\

Step 3: Solve for x

The equation is now a quadratic equation. It can be simplified by expanding the bracket to obtain: \\(x^2 + 7x + 12 = 6\)\\Then, rearrange the equation to have all terms on one side, which helps in further factorization: \\(x^2 + 7x + 6 = 0\)\\Now, the equation can be factored: \\(x^2 + 7x + 6 = 0\) factoring this gives \\(x+1)(x+6)=0\)\\Hence the solutions of the equation are \\x = -1 and x = -6\)\\

Step 4: Check for Extraneous Solutions

In logarithmic equations, it's important to check the solutions, since not all values that make the equation true are necessarily in the domain of the original logarithmic expressions. In this case, both x = -1, and x = -6 are not in the domain of the original logarithmic expression, because the arguments of the logarithms, (x+3) and (x+4), would be non-positive. Hence, there is no solution to this equation.

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is crucial when solving logarithmic equations. One essential property states that the sum of two logarithms with the same base can be combined into a single logarithm by multiplying their arguments: \(\text{log}_a(m) + \text{log}_a(n) = \text{log}_a(mn)\). This allows us to simplify complex expressions and prepare them for conversion to exponential form. Here's a quick list of some other properties that often come in handy:

• Product Property: \(\text{log}_a(mn) = \text{log}_a(m) + \text{log}_a(n)\)
• Quotient Property: \(\text{log}_a(\frac{m}{n}) = \text{log}_a(m) - \text{log}_a(n)\)
• Power Property: \(\text{log}_a(m^n) = n \times \text{log}_a(m)\)

These properties enable us to manipulate logarithmic equations effectively, setting the stage for solving more complex problems.

Converting Logarithms to Exponentials

To solve logarithmic equations, we often convert them into their exponential counterparts, which makes the problem more straightforward to solve. The conversion relies on understanding that if \(\text{log}_a(b) = c\), then the exponential form is \(a^c = b\). This transformation moves us from the logarithm's world into the more familiar realm of exponents.

In the original exercise, once the logarithmic properties are applied, we get \(\text{log}_6((x+3)(x+4)) = 1\). Applying the conversion to an exponential equation gives us \(6^1 = (x+3)(x+4)\), transforming the problem into a basic algebraic equation which we are better equipped to solve.

Solving Quadratic Equations

When logarithmic equations are transformed, they often become quadratic equations. A quadratic equation has the standard form \(ax^2 + bx + c = 0\), and we can solve for \(x\) using several methods, including factoring, completing the square, or applying the quadratic formula. In our example, after expanding and rearranging terms, the equation \(x^2 + 7x + 12 = 6\) simplifies to the quadratic equation \(x^2 + 7x + 6 = 0\).

Factoring is one way to find solutions, which involves expressing the quadratic in the form \((x+p)(x+q)=0\), where \(p\) and \(q\) are the solutions. This way, we find our potential solutions, which leads us to the next crucial step: checking for extraneous solutions.

Checking for Extraneous Solutions

In the context of logarithmic equations, not all solutions obtained algebraically are valid. Some may be extraneous, which means they don't satisfy the original equation. It's essential to check our solutions against the domain of the logarithmic function, as logarithms are only defined for positive arguments. In the exercise, we obtained the potential solutions \(x = -1\) and \(x = -6\), but when these values are plugged back into the original logarithmic expressions, they resulted in a non-positive inside the log function, which is not permissible.

By checking for extraneous solutions, we assure that our final answer is within the domain of the logarithm, maintaining the integrity of our solutions. Skipping this step can lead to incorrect answers and misunderstandings in the properties and behavior of logarithmic functions.