Problem 90 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

For part (a), solutions are \( x = 2 \) and \( x = -1 \); for part (b), solutions are \( x = 3 \) and \( x = -2 \).

Step 1: Recognize the Logarithm Definition

For both parts (a) and (b), start by recalling the definition of a logarithm: if \( \log_b(a) = c \), then \( b^c = a \). This will help us rewrite the logarithmic equations in exponential form.

Step 2: Convert to Exponential Form for Part (a)

Given \( \log_{2}(x^{2}-x) = 1 \), apply the definition of logarithms: \( 2^1 = x^2 - x \), simplifying to \( x^2 - x = 2 \).

Step 3: Solve the Quadratic Equation for Part (a)

Rearrange the equation \( x^2 - x - 2 = 0 \). Factor this quadratic to \( (x - 2)(x + 1) = 0 \). Thus, the solutions are \( x = 2 \) and \( x = -1 \).

Step 4: Verify Solutions for Part (a)

Substitute \( x = 2 \) and \( x = -1 \) back into the original equation to ensure validity: both solutions make \( x^2 - x \) positive, so they are valid.

Step 5: Convert to Exponential Form for Part (b)

For \( \log_{6}(x^{2}-x) = 1 \), convert using the definition: \( 6^1 = x^2 - x \), simplifying to \( x^2 - x = 6 \).

Step 6: Solve the Quadratic Equation for Part (b)

Rearrange the equation \( x^2 - x - 6 = 0 \). Factor this quadratic as \( (x - 3)(x + 2) = 0 \). Thus, the solutions are \( x = 3 \) and \( x = -2 \).

Step 7: Verify Solutions for Part (b)

Substitute \( x = 3 \) and \( x = -2 \) back into the original equation to ensure validity: both solutions make \( x^2 - x \) positive, so they are valid.

Key Concepts

Logarithmic Equations

Logarithmic equations are equations that involve logarithms, which are mathematical expressions representing how many times a number, called the base, must be multiplied by itself to produce another number. In most algebra exercises, you might see something like \( \log_b(a) = c \), meaning the base \( b \) raised to power \( c \) gives the number \( a \).

Understanding how to manipulate these equations is crucial, especially in Intermediate Algebra, because it allows us to solve for unknown variables by converting equations into exponential form.

• Always remember the conversion: If \( \log_b(a) = c \), then \( b^c = a \).
• The logarithm function is the inverse of the exponential function.
• Knowing how to switch between exponential and logarithmic forms can make complex equations simpler.

In our original exercise, we started with equations like \( \log_2(x^2 - x) = 1 \). By applying the definition, we rearrange to exponential form: \( 2^1 = x^2 - x \). This pivotal step simplifies our equation and sets the stage for solving it effectively.

Quadratic Equations

Quadratic equations are a key element in algebra and typically take the form \( ax^2 + bx + c = 0 \). In the exercises provided, after converting the logarithmic equation to an exponential form, we found ourselves with quadratic equations. For instance, \( x^2 - x - 2 = 0 \) or \( x^2 - x - 6 = 0 \).

Solving quadratic equations generally involves one of these methods:

• Factoring: This method requires expressing the quadratic in the form \((x - p)(x - q) = 0\), where \( p \) and \( q \) are solutions.
• Completing the square: Often useful for quadratics that are not easily factored.
• Quadratic formula: Given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), applicable to any quadratic equation.

Using factoring, we can solve the equations from the exercise. For example, \( x^2 - x - 2 = (x - 2)(x + 1) = 0 \) leads us to \( x = 2 \) and \( x = -1 \) as solutions. It's important to verify each solution by plugging it back into the original equation to ensure that it satisfies the condition set by the logarithm.

Exponential Form

In algebra, exponential form is a way of writing an expression that uses a constant base raised to a variable exponent. It is the key to resolving logarithmic equations that have been introduced into exponential form. When you see an equation like \( b = a^c \), you know this is in exponential form.

Converting a logarithmic equation into exponential form is a powerful strategy, simplifying our work significantly. Here's a simple guide to performing this conversion:

• Begin with the logarithmic equation \( \log_b(a) = c \).
• Apply the conversion: shift to exponential form by expressing it as \( a = b^c \).
• This elucidation helps us apply subsequent techniques like solving quadratic equations.

For example, by rewriting \( \log_{6}(x^2 - x) = 1 \) as \( 6^1 = x^2 - x \), we simplify our process, making it possible to handle the expression using straightforward algebraic techniques. Knowing how to change forms fluently is a skill that will enhance your problem-solving abilities in mathematics.