Problem 22 - Algebra A Combined Function

Question

Solve each equation. $$ \log _{6}\left(x^{2}-x\right)=1 $$

Step-by-Step Solution

Answer

The solutions are \( x = 3 \) and \( x = -2 \).

Step 1: Understand the Logarithmic Equation

The given equation is \( \log _{6}\left(x^{2}-x\right)=1 \). This means that the logarithm of \( x^2 - x \) with base 6 is equal to 1.

Step 2: Convert to Exponential Form

Use the definition of a logarithm to convert the equation from logarithmic to exponential form. If \( \log_b(a) = c \), then \( b^c = a \). Therefore, \( 6^1 = x^2 - x \).

Step 3: Simplify the Equation

From Step 2, we have \( 6 = x^2 - x \). Rearrange it to represent a quadratic equation: \( x^2 - x - 6 = 0 \).

Step 4: Factoring the Quadratic Equation

Factor the quadratic equation \( x^2 - x - 6 = 0 \). This can be done by finding two numbers that multiply to -6 and add to -1. The numbers are 2 and -3. So, it factors to \((x - 3)(x + 2) = 0 \).

Step 5: Solve for x

Set each factor equal to zero: \( x - 3 = 0 \) and \( x + 2 = 0 \). Solve these equations to find the solutions for \( x \): \( x = 3 \) and \( x = -2 \).

Step 6: Verify Solutions

Check each solution to ensure they make the original logarithm defined and valid. For \( x = 3 \), substitute back into \( x^2 - x = 9 - 3 = 6 \), which is greater than zero and valid. For \( x = -2 \), substitute back into \( x^2 - x = 4 + 2 = 6 \), which is also greater than zero and valid.

Key Concepts

Exponential Form

When you encounter a logarithmic equation, it's often helpful to rewrite it in its exponential form. This can make the equation easier to solve or manipulate.
In our example, we began with the equation \(\log _{6}(x^{2}-x) = 1.\)By using the definition of a logarithm, which states that \(\log_b(a) = c\) means \(b^c = a\), we can convert this into exponential form:

• The base \(b\) is 6,
• The exponent \(c\) is 1,
• The result \(a\) becomes \(x^2 - x\).

Thus, the equation becomes \(6^1 = x^2 - x\). This transformation shows that \(x^2 - x\) equals 6, providing a simpler equation to work with. This step is crucial because exponential equations are often easier to handle than their logarithmic counterparts.

Quadratic Equations

After converting the log equation into exponential form, we receive a quadratic equation: \(x^2 - x - 6 = 0\). Quadratic equations are polynomials with degree 2 and they often take the form \(ax^2 + bx + c = 0\).
To solve a quadratic equation, one effective method is factoring. In the context of our equation, we need two numbers that multiply to the constant term (-6) and add up to the linear coefficient (-1).

• Those two numbers are 2 and -3.
• This allows the expression \(x^2 - x - 6\) to be factored as \((x - 3)(x + 2) = 0\).

Once factored, it's straightforward to solve for \(x\):

• Set each factor to zero, giving \(x - 3 = 0\) and \(x + 2 = 0\).
• Solving these gives us \(x = 3\) or \(x = -2\).

These solutions indicate where the quadratic equation will intersect the x-axis.

Solutions Verification

Even after solving a quadratic equation, it's important to verify the possible solutions in the context of the original logarithmic equation.
Substitute each solution back into the expression inside the logarithm to ensure it makes the log valid (i.e., the argument of the log must be positive):

• For \(x = 3\), substituting into \(x^2 - x\) gives \(9 - 3 = 6\). The result is positive, indicating that the logarithm is defined for this solution.
• For \(x = -2\), substituting gives \(4 + 2 = 6\), which is also positive. Therefore, \(x = -2\) also results in a valid log argument.

It's crucial to verify solutions because if the argument inside the logarithm is less than or equal to zero, the logarithm would be undefined. Ensuring all solutions make all parts of the original equation valid saves from mathematical pitfalls and ensures accuracy.