Problem 12 - Algebra 2 and Trigonometry

Question

Solve each equation for the variable and check. \(\log x+\log (x-1)=\log 12\)

Step-by-Step Solution

Answer

The solution is \( x = 4 \). Negative solution is invalid as log of a negative number is undefined.

Step 1: Apply the Product Rule of Logarithms

The equation is given as \( \log x + \log (x-1) = \log 12 \). According to the product rule of logarithms, \( \log a + \log b = \log (ab) \). Apply this rule to the left side of the equation: \( \log (x(x-1)) = \log 12 \).

Step 2: Remove the Logarithms

Since the equation \( \log (x(x-1)) = \log 12 \) involves logarithms on both sides, if two logs with the same base are equal, then their arguments must be equal. Therefore, set the arguments equal to each other: \( x(x-1) = 12 \).

Step 3: Expand and Rearrange the Equation

Expand the left-side expression: \( x(x-1) = x^2 - x \). Replace the equation to \( x^2 - x = 12 \). Rearrange it to form a quadratic equation: \( x^2 - x - 12 = 0 \).

Step 4: Factor the Quadratic Equation

Factor the quadratic equation \( x^2 - x - 12 = 0 \). The factor pairs of \(-12\) are \( (3, -4) \) because their sum is \(-1\). Thus, factor the equation to \( (x-4)(x+3) = 0 \).

Step 5: Solve for x

Set each factor equal to zero: \( x-4 = 0 \) or \( x+3 = 0 \). Solve each equation to find the possible solutions: \( x = 4 \) and \( x = -3 \).

Step 6: Check Both Solutions

Substitute \( x = 4 \) into the original equation: \( \log 4 + \log (4-1) = \log 12 \). Simplify: \( \log 4 + \log 3 = \log 12 \). This is correct by the product rule. Now check \( x = -3 \): substitute \( \log (-3) + \log (-3-1) \), which is undefined since logarithms of negative numbers are not real. Thus, \( x = -3 \) is not a valid solution.

Key Concepts

Product Rule of Logarithms

When you encounter logarithmic expressions like \( \log a + \log b \), there's a handy tool called the Product Rule of Logarithms. This rule tells us that the sum of two logarithms is equal to the logarithm of the product of their arguments. In simple terms:

• \( \log a + \log b = \log (ab) \)

This rule allows us to combine the logarithms, simplifying our equations. For example, \( \log x + \log (x-1) = \log(x(x-1)) \) becomes \( \log x(x-1) \). This conversion is essential since it helps us move forward to further simplify and solve equations. Notice how much cleaner and more manageable the equation becomes once we apply this rule. Breaking down logarithmic expressions in this way is crucial in solving complex logarithmic equations efficiently.

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). In our exercise, we derived the quadratic equation \( x^2 - x - 12 = 0 \) by expanding and rearranging \( x(x-1) = 12 \). Some important characteristics of quadratic equations include:

• They have a degree of 2, meaning they involve \( x^2 \).
• The graph of a quadratic equation is a parabola.
• They may have two real roots, one real root, or complex roots based on the discriminant value.

Understanding and recognizing the structure of quadratics assist in determining solution methods. Furthermore, mastering simple rearrangements can allow us newcomers to solve these equations confidently. In practice, this generally means isolating the squared term and forming a recognizable equation to work with.

Arguments Equality

In logarithmic equations, achieving arguments equality is fundamental. If the logarithms of two expressions are equal, the expressions themselves must also be equal. This principle works because logarithmic functions are one-to-one. Therefore, when we have \( \log (x(x-1)) = \log 12 \), we can safely equate their arguments:

• \( x(x-1) = 12 \)

This step is critical because it removes the logarithms, leaving us with a simpler algebraic equation to solve. Knowing this property is particularly useful as it allows us to shift from dealing with potentially cumbersome logarithms to simpler algebraic forms, thus enabling us to explore solutions using algebraic methods like expansion and factoring.

Factoring Quadratics

Factoring is a classic and effective method for solving quadratic equations like \( x^2 - x - 12 = 0 \). To factor, we search for two numbers whose product is the constant term and whose sum is the coefficient of the linear term. In our equation:

• The product of the numbers should be −12 (the constant term).
• Their sum should be −1 (the coefficient of \( x \)).

After testing factor pairs of −12, we identify 3 and −4 as suitable numbers because they satisfy both conditions:

• 3 and −4 multiply to give −12.
• 3 plus −4 equals −1.

Thus, we can factor the quadratic into \( (x - 4)(x + 3) = 0 \). Factoring makes it easier to apply the zero-product property, simplifying the process of finding the solutions to \( x \). This method can transform a daunting quadratic into straightforward linear equations that students can solve easily.