Problem 68 - 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. $$ \log _{6}(x+5)+\log _{6} x=2 $$

Step-by-Step Solution

Answer

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

Step 1: Combine the Logarithms

Apply the logarithmic property \(\log a + \log b = \log ab\). So the given equation \(\log _{6}(x+5) + \log _{6} x = 2\) becomes \(\log _{6} ((x+5)x) = 2\).

Step 2: Apply Anti-Logarithm

To get rid of the logarithm, apply an anti-logarithm base 6 on both sides which gives us the equation \(6^{\log _{6}((x+5)x)} = 6^2\). It simplifies to \((x^2 + 5x) = 36\).

Step 3: Solve the Quadratic

Solving the quadratic equation \(x^2 + 5x - 36 = 0\) will lead to \((x - 4)(x + 9) = 0\), which gives us potential solutions \(x = 4\) and \(x = -9\).

Step 4: Check the Domain

Logarithms are undefined for negative values, so validating potential solutions against the original equation is necessary. In the original ,negative values are not accepted because one cannot take a logarithm of a negative number. Therefore, \(x = -9\) is not valid, and the only solution is \(x = 4\).

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is like unlocking a language to simplify complex expressions. In our exercise, we use the property of logarithms that states \(\log a + \log b = \log(ab)\). This allows us to combine two logarithms with the same base into a single logarithm.

Here's how it works: if you have \(\log_6(x+5) + \log_6 x\), you can combine these using the product property to get \(\log_6((x+5) \cdot x)\). This transforms our equation into a simpler form, helping you proceed toward the solution more efficiently.

Utilizing these properties is crucial in solving logarithmic equations and is a fundamental skill in algebra. Remember, simplifying expressions is often the key to solving equations.

Domain of a Function

When dealing with logarithms, understanding the domain is critical. The domain of a function includes all the allowable values of \(x\) that make the function defined. For logarithmic functions, the argument must be greater than zero.

In our equation \(\log_6(x+5) + \log_6 x = 2\), we need both \(x+5\) and \(x\) to be greater than zero:

• \(x > 0\)
• \(x + 5 > 0\)

This simplifies to \(x > 0\).

It's important to always check potential solutions against this domain. In this case, only \(x = 4\) satisfies the domain, whereas \(x = -9\) does not because it would result in negative arguments, which are not valid for logarithms.

Quadratic Equations

Once the logarithms are combined and the equation transformed, we end up with a quadratic equation. Quadratic equations take the standard form \(ax^2 + bx + c = 0\). Solving these can be done through factoring, completing the square, or using the quadratic formula.

In our problem, we solve \(x^2 + 5x - 36 = 0\) by factoring:

• \((x - 4)(x + 9) = 0\)

By setting each factor equal to zero, we find the potential solutions:

• \(x = 4\)
• \(x = -9\)

Factoring is often the easiest method for solving a quadratic equation when the factors are accessible and integers.

Exact and Approximate Solutions

Once you solve the equation, you need to determine which solutions are valid. An exact solution is derived directly from algebraic manipulation without approximation. For instance, \(x = 4\) and \(x = -9\) are exact solutions.

However, not all exact solutions are necessarily valid in the context of the problem. We previously determined the domain and rejected \(x = -9\) because it falls outside the domain of the function. Therefore, \(x = 4\) is the valid solution.

Approximate solutions come into play when exact solutions are complex or irrational. In such cases, a calculator might be used to find a decimal approximation. This doesn't apply directly here, as \(x = 4\) is already exact and valid, but approximations are crucial in many real-world applications.