Problem 43 - Precalculus : Building Concepts and Connections

Question

Solve the logarithmic equation and eliminate any extraneous solutions. If there are no solutions, so state. $$\log (x+1)+\log (x-1)=0$$

Step-by-Step Solution

Answer

The solution to the given logarithmic equation is \(x = 1\).

Step 1: Combine the Logarithms

When the logs have the same base and are being added, it can be simplified to take the log of the product of their arguments. Meaning, \(\log (x+1)+\log (x-1) = \log [(x+1)(x-1)]\) .

Step 2: Simplify the Equation

Now the simplified logarithm can be changed to an exponential form. This will allow us to solve the equation. This gives us \((x+1)(x-1) = 1^0\) .

Step 3: Solve for x

Now, we expand the left-side of equation and simplify. This gives us \(x^2 - 1 = 0\). Solving for \(x\) gives \(x = ±1\).

Step 4: Check for Extraneous solutions

It is important to check the solutions in the original equation since logarithm of zero or a negative number is undefined. Check if \(x = 1\) and \(x = -1\) are solutions. We see that when \(x = -1\), one of the terms of the equation becomes undefined. Therefore, the correct solution is \(x =1\) .

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is crucial when solving logarithmic equations. A logarithm is the power to which a number (the base) must be raised to obtain another number. For example, if we have the equation \(\log_b(x) = y\), this can be interpreted as \(b^y = x\). When solving equations involving logarithms, there are several key properties we use:

• Product Property: The log of a product is the sum of the logs of the factors, which we use in Step 1 of the exercise. Mathematically, \(\log_b(mn) = \log_b(m) + \log_b(n)\).
• Quotient Property: The logarithm of a quotient is the difference of the logarithms. That is, \(\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)\).
• Power Property: \(\log_b(m^n) = n\log_b(m)\), which tells us that a logarithm of a power allows the exponent to be brought in front as a multiplier.

Recognizing and applying these properties properly is essential for simplifying and solving logarithmic equations efficiently.

Exponential Form

Transforming a logarithmic equation into its exponential form simplifies the process of finding the variable. This conversion is based on the definition of a logarithm. The equation \(\log_b(x) = y\) can be rewritten as \(b^y = x\), where \(b\) is the base of the logarithm, \(y\) is the logarithm of the number \(x\), and \(x\) is the result when \(b\) is raised to the power of \(y\).

In Step 2 of our exercise, we use this principle to write \(\log [(x+1)(x-1)] = 0\) as \(1^0 = (x+1)(x-1)\). Since any number raised to the power of zero is 1, it simplifies the equation to a form where we can easily solve for \(x\) by expanding and rearranging terms.

Extraneous Solutions

An extraneous solution is a result obtained during the process of solving an equation that is not a valid solution to the original equation. In logarithmic equations, extraneous solutions often arise when the logarithm of a negative number or zero occurs, which is undefined, as logarithms are only defined for positive numbers.

In Step 4 of our solution, we must verify the potential solutions \(x = ±1\) in the original equation. The term \(\log (x-1)\) becomes undefined when \(x = -1\), as the argument of the logarithm must be positive. Thus, \(x = -1\) is an extraneous solution and is discarded, leaving us with the valid solution \(x = 1\). It is always important to check for extraneous solutions to ensure that all solutions satisfy the original equation.