Problem 25 - Algebra for College Students

Question

Solve each logarithmic equation and express irrational solutions in lowest radical form. $$ \log (x+1)=\log 3-\log (2 x-1) $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{-1 + \sqrt{33}}{4} \).

Step 1: Apply Logarithm Properties

We start by applying the properties of logarithms. Recall that \( \log a - \log b = \log \frac{a}{b} \). Therefore, rewrite the equation:\[ \log(x + 1) = \log \frac{3}{2x - 1} \]

Step 2: Equate the Arguments

Since the logarithms on both sides of the equation are equal, their arguments must be equal as well. Thus,:\[ x + 1 = \frac{3}{2x - 1} \]

Step 3: Cross Multiply

To eliminate the fraction, cross-multiply:\[ (x + 1)(2x - 1) = 3 \]This equation results in:\[ 2x^2 - x + 2x - 1 = 3 \]

Step 4: Simplify and Rearrange

Simplify the equation by combining like terms and moving all terms to one side:\[ 2x^2 + x - 1 - 3 = 0 \]\[ 2x^2 + x - 4 = 0 \]

Step 5: Use the Quadratic Formula

The quadratic equation is of the form \( ax^2 + bx + c = 0 \), where \(a = 2\), \(b = 1\), and \(c = -4\). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we have:\[ x = \frac{-1 \pm \sqrt{1 + 32}}{4} \]\[ x = \frac{-1 \pm \sqrt{33}}{4} \]

Step 6: Determine Which Solution Is Valid

We need to ensure the solution value makes the original logarithmic expression valid (i.e., arguments of log > 0).1. For \( x = \frac{-1 + \sqrt{33}}{4} \), check: - \( x + 1 = \frac{3 + \sqrt{33}}{4} > 0 \) - \( 2x - 1 = \frac{-3 + 2\sqrt{33}}{4} > 0 \) 2. The negative root \( x = \frac{-1 - \sqrt{33}}{4} \) does not satisfy these conditions.

Key Concepts

Properties of Logarithms

When solving logarithmic equations, understanding the properties of logarithms is essential. These properties allow us to simplify and manipulate logarithmic expressions effectively. Here are some key properties of logarithms that are often used:

• Product Property: \( \log_b(MN) = \log_b M + \log_b N \)
• Quotient Property: \( \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N \)
• Power Property: \( \log_b(M^p) = p \cdot \log_b M \)

The original problem uses the quotient property to simplify the equation: \[ \log(x + 1) = \log\left(\frac{3}{2x - 1}\right) \]This property tells us that if we subtract the logarithm of one quantity from another, it is equivalent to taking the logarithm of their quotient. This allows us to set the equations equal and solve further.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In the exercise, after rearranging and simplifying the equation from the logarithmic expressions, we acquired the quadratic equation:\[ 2x^2 + x - 4 = 0 \]Here, the coefficients are: \( a = 2 \), \( b = 1 \), and \( c = -4 \). These are substituted into the quadratic formula to find the values of \( x \). The solutions involve calculating the discriminant \( b^2 - 4ac \), which in this case is \( 1 + 32 = 33 \), leading us to solutions involving a square root.

Solving Equations

Solving equations involves finding the values of the variable that satisfy the given mathematical condition. With logarithmic equations, once the logs are eliminated using their properties, you may transform the equation into a more recognizable form, often requiring expressions like cross-multiplication. Here's a simple guide:

• Isolate the logarithmic terms if possible.
• Apply the properties of logarithms to simplify.
• Set the logarithmic expressions equal if they have the same base.
• Solve for the variable by conducting algebraic operations such as cross-multiplying.

Cross-multiplication was used in the exercise to clear the fractional equation, allowing simplification into a quadratic form, which was then resolved using the quadratic formula.

Radical Expression

Radical expressions involve roots, such as square roots. When solving equations, especially quadratic ones, the solutions sometimes involve radicals. In the exercise, the solution to the quadratic equation resulted in:\[ x = \frac{-1 \pm \sqrt{33}}{4} \]The term \( \sqrt{33} \) is a radical expression. When dealing with radicals in solutions, it's crucial to ensure they are simplified to their lowest form and that they fit the context of the problem. For logarithmic equations, we also check that these radicals result in expressions that are defined over the replacements back in the original context, ensuring no negative or undefined logarithm arguments.