Problem 90 - Elementary and Intermediate Algebra

Question

a. \(\log _{2}\left(x^{2}-x\right)=1\) b. \(\log _{6}\left(x^{2}-x\right)=1\)

Step-by-Step Solution

Answer

a) \(x = 2\); b) \(x = 3\)

Step 1: Understand the given equation

We have two different logarithmic equations to solve, each with the form \(\log_b(y) = 1\). For part (a), the base is 2, and for part (b), the base is 6. Our goal is to solve each equation for \(x\).

Step 2: Rewrite the logarithmic equation into exponential form (Part a)

For the equation \(\log_2(x^2 - x) = 1\), use the property of logarithms \(\log_b(y) = z\) implies \(y = b^z\). So, \(x^2 - x = 2^1\).

Step 3: Solve the equation from Part a

We now have \(x^2 - x = 2\). Rearrange the equation into standard quadratic form: \(x^2 - x - 2 = 0\). Factor the quadratic equation: \((x - 2)(x + 1) = 0\). Therefore, the solutions are \(x = 2\) and \(x = -1\).

Step 4: Check feasibility for Part a

Substitute \(x = 2\) and \(x = -1\) into the original expression \(x^2 - x\) to ensure the value is positive, as logarithms are only defined for positive arguments. For \(x = 2\), we get 2 and for \(x = -1\), we get -2. Thus, the valid solution is \(x = 2\).

Step 5: Rewrite the logarithmic equation into exponential form (Part b)

For the equation \(\log_6(x^2 - x) = 1\), convert it to exponential form: \(x^2 - x = 6^1\).

Step 6: Solve the equation from Part b

Simplify to \(x^2 - x = 6\). Rearrange into \(x^2 - x - 6 = 0\). Factor the quadratic equation: \((x - 3)(x + 2) = 0\). Therefore, the solutions are \(x = 3\) and \(x = -2\).

Step 7: Check feasibility for Part b

Substitute \(x = 3\) and \(x = -2\) into \(x^2 - x\) to check for positivity. For \(x = 3\) the result is 6 and for \(x = -2\), the result is -6. Therefore, the valid solution is \(x = 3\).

Key Concepts

Understanding Logarithmic Functions

Logarithmic functions are an essential mathematical concept used to solve various types of equations, especially exponential equations. They are the inverse of exponential functions and are used to find the exponent to which a base must be raised to obtain a certain number. In essence, if you have a logarithmic equation like \( \log_b(y) = z \), it means that \( b^z = y \).

The base \( b \) is a crucial component of logarithmic functions. It determines the growth rate of the logarithm. For example, in our original exercise, we have two bases, 2 and 6, indicating how the corresponding solutions will differ due to their base's strength.

Logarithmic functions have specific properties and rules that help in simplifying and solving equations. For instance:

• \( \log_b(xy) = \log_b(x) + \log_b(y) \)
• \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \)
• \( \log_b(x^n) = n\log_b(x) \)

Understanding these properties can greatly assist in transforming complex logarithmic equations into simpler forms.

Converting to Exponential Form

To solve a logarithmic equation, it is often converted into its exponential form. This is achieved using the fundamental property that states if \( \log_b(y) = z \), then \( y = b^z \).

In practical terms, this means taking a logarithmic equation and rewriting it in a form that is typically easier to solve. For example, in the original exercise, converting \( \log_2(x^2-x) = 1 \) into exponential form results in \( x^2-x = 2^1 \). This format is straightforward and sets up the equation nicely for further algebraic manipulation. The same process applies to \( \log_6(x^2-x) = 1 \), leading to \( x^2-x = 6^1 \).

Understanding the exponential form decentralizes the task by removing the logarithm, making the equation more approachable, whether it boils down to basic algebra or more complex procedures.

Solving Quadratic Equations

Once a logarithmic equation is converted into exponential form, solving it often leads to the necessity of solving a quadratic equation. Quadratic equations usually take the standard form of \( ax^2 + bx + c = 0 \). Solving them can be done through various methods:

• **Factoring**: This method is often the simplest, provided the equation is factorable. For example, \( x^2-x-2=0 \) can be factored to \((x-2)(x+1)=0\).
• **Quadratic Formula**: When factoring isn't straightforward, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) can be used to find solutions.
• **Completing the Square**: This method involves manipulating the equation to make one side a perfect square trinomial, though it's less commonly used for straightforward problems.

Quadratic equations can also have multiple solutions, which necessitates checking the feasibility of each in the context of the original equation. As we see in the original exercise, some solutions may not be valid, thus ensuring they fit within the constraints of the problem, such as requiring logarithms to have positive arguments, is essential.