Problem 19 - Intermediate Algebra

Question

For Problems \(15-22\), solve each logarithmic equation. $$ \log (x+2)-\log (2 x+1)=\log x $$

Step-by-Step Solution

Answer

The solution is \(x = 1\).

Step 1: Use Logarithm Properties

The given equation is \( \log (x+2) - \log (2x+1) = \log x \). Utilizing the property of logarithms that \( \log a - \log b = \log \frac{a}{b} \), the left side can be rewritten as \( \log \frac{x+2}{2x+1} = \log x \).

Step 2: Equate Arguments of Logarithms

Since the logarithms have the same base and are equal, their arguments must be equal. This gives us the equation: \( \frac{x+2}{2x+1} = x \).

Step 3: Clear the Fraction

Multiply both sides of the equation by \( 2x + 1 \) to clear the fraction: \((x+2) = x(2x+1)\).

Step 4: Simplify and Rearrange

Distribute and rearrange the equation: \( x + 2 = 2x^2 + x \). Then simplify to \( 0 = 2x^2 + x - x - 2 \) which becomes \( 2x^2 - 2 = 0 \).

Step 5: Factor the Quadratic Equation

Factor the quadratic equation \( 2x^2 - 2 = 0 \). This can be rewritten as \( 2(x^2 - 1) = 0 \), which factors further to \((x-1)(x+1) = 0 \).

Step 6: Solve for x

Set each factor equal to zero: \(x - 1 = 0\) and \(x + 1 = 0\). Solving these gives \(x = 1\) and \(x = -1\).

Step 7: Validate Solutions

Since logarithms are only defined for positive values, \(x = -1\) is not valid as it makes \(\log x\) undefined. Thus, \(x = 1\) is the only valid solution.

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is crucial for solving logarithmic equations. One of the fundamental properties is the product, quotient, and power rules. Here we focus on the quotient property:

• The quotient rule states: \( \log a - \log b = \log \frac{a}{b} \).
• This property allows transformation from a subtraction of two logarithms into a single logarithm of a quotient.

To use this, look at the equation \( \log (x+2) - \log (2x+1) = \log x \). By applying the quotient rule, this simplifies to \( \log \frac{x+2}{2x+1} = \log x \). This transformation makes it easier to isolate the variable by reducing the complexity of the equation, letting us focus on the arguments of the logarithms and solve them directly.

Solving Equations

Once logarithm properties are applied, the next step is solving the equation derived from their arguments. When two logarithms with the same base are equal, their arguments must also be equal. This simplifies the problem:

• Given: \( \log \frac{x+2}{2x+1} = \log x \).
• Equate the arguments: \( \frac{x+2}{2x+1} = x \).

With the logarithms eliminated, focus on solving this rational equation. The key is to clear the fraction by multiplying both sides by \(2x + 1\), leading to: \(x + 2 = x(2x + 1)\). This results in a standard algebraic equation. Solving this involves simplifying and rearranging terms, a critical step before addressing any quadratic component that arises.

Quadratic Equations

Reaching a quadratic equation often happens after manipulating expressions. In our exercise, we derived a quadratic from simplifying the equation: \( x + 2 = 2x^2 + x \).

• Rearrange to standard form: \( 0 = 2x^2 + x - x - 2 \), or simplified \( 2x^2 - 2 = 0 \).
• Factor the quadratic: \( 2(x^2 - 1) = 0 \).

The quadratic \( x^2 - 1 \) factors to \((x-1)(x+1)\). Solving these, we find potential solutions: \( x = 1 \) and \( x = -1 \).However, not all solutions may fit the original equation context due to the domain restrictions of logarithms, as they are undefined for non-positive numbers. Thus, checking validity gives only \( x = 1 \) as acceptable, since \( x = -1 \) is not in the valid domain for logarithms.