Problem 26 - Algebra for College Students

Question

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

Step-by-Step Solution

Answer

The solution is \( x = \frac{-1 + \sqrt{65}}{2} \).

Step 1: Eliminate the Logs

We start by rewriting the equation in terms of the same logarithmic base. Initially, we have \( \log(x - 2) = 1 - \log(x + 3) \). Add \( \log(x + 3) \) to both sides to get: \( \log(x - 2) + \log(x + 3) = 1 \).

Step 2: Use Logarithmic Properties

Apply the property of logarithms that combines sums of logs into a single log: \( \log(a) + \log(b) = \log(a \cdot b) \). This results in the equation: \( \log((x - 2)(x + 3)) = 1 \).

Step 3: Convert Logarithmic Equation to Exponential Form

By converting the log equation \( \log((x - 2)(x + 3)) = 1 \) to its exponential form, we get \( 10^1 = (x - 2)(x + 3) \). This simplifies to \( 10 = (x - 2)(x + 3) \).

Step 4: Expand and Simplify the Quadratic Equation

Expand the equation from Step 3: \( 10 = x^2 + 3x - 2x - 6 \), which simplifies to \( 10 = x^2 + x - 6 \). Rearrange this to form a standard quadratic equation: \( x^2 + x - 16 = 0 \).

Step 5: Solve the Quadratic Equation

Solve \( x^2 + x - 16 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = -16 \). Calculate the discriminant: \( b^2 - 4ac = 1 + 64 = 65 \). Therefore, \( x = \frac{-1 \pm \sqrt{65}}{2} \).

Step 6: Validate the Solutions

Verify that the values \( x = \frac{-1 + \sqrt{65}}{2} \) and \( x = \frac{-1 - \sqrt{65}}{2} \) satisfy the original condition \( x > 2 \) (because \( x - 2 \) needs to be positive for the log). Calculating, only \( x = \frac{-1 + \sqrt{65}}{2} \) is greater than 2.

Key Concepts

Properties of Logarithms

Logarithms have several properties that make them useful in solving equations, particularly when you need to combine or simplify expressions. One of the crucial properties that helps in solving equations is:

• The Product Property: This states that the sum of two logs with the same base is equivalent to the logarithm of the product of their arguments: \( \log(a) + \log(b) = \log(a \cdot b) \).

This property is especially useful when you encounter the sum of logarithms and wish to simplify it into a single log expression.
In the exercise, we applied this property to combine \( \log(x - 2) + \log(x + 3) \) into \( \log((x - 2) \cdot (x + 3)) \). This step is pivotal as it allows us to transition to the exponential form and solve the equation more straightforwardly.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The general solution is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula provides the roots of any quadratic equation if the equation does indeed have real solutions.
In our example, after manipulation and rearrangement, the equation \( x^2 + x - 16 = 0 \) was obtained. By identifying \( a = 1 \), \( b = 1 \), and \( c = -16 \), we used the quadratic formula to find the roots.
This method ensures that even when faced with complex numbers, you can retrieve solutions effectively, showcasing the formula's universality.

Exponential Form

Changing a logarithmic equation to exponential form is a fundamental step when solving such equations. It involves leveraging the definition of a logarithm:

• If \( \log_b(a) = c \), then in exponential form it is \( b^c = a \).

This forms the basis for converting logs into a more manageable algebraic form.
In our solution, we converted \( \log((x - 2)(x + 3)) = 1 \) to its exponential form, which is \( 10^1 = (x - 2)(x + 3) \). This conversion simplifies the equation to direct multiplication rather than dealing with logarithms.
Applying the exponential form helps to bridge the gap between logarithmic expressions and conventional algebra, aiding in simplifying and ultimately solving the problem.

Solving Quadratic Equations

Once a quadratic equation has been established, solving it is typically the next step. Apart from the quadratic formula, other methods include:

• Factoring, where the equation is expressed as a product of its roots.
• Completing the square, transforming the expression to a perfect square form.

In many cases, however, the quadratic formula is preferred due to its straightforward applicability.
For the quadratic \( x^2 + x - 16 = 0 \), we used the formula, which is especially beneficial when the equation doesn't easily factor or complete the square. Calculating the roots gave us \( x = \frac{-1 \pm \sqrt{65}}{2} \).
This solution method is robust across various types of quadratic equations, and choosing the right method depends on the specific characteristics of the equation at hand.