Problem 141 $$ \log _{2}\left(\frac{x-2}{x... [FREE SOLUTION]

Chapter 7: Problem 141

$$ \log _{2}\left(\frac{x-2}{x-1}\right)-1=\log _{2}\left(\frac{3 x-7}{3 x-1}\right) $$

Short Answer

Expert verified

The short version of the answer is: To solve the logarithmic equation $\log _{2}\left(\frac{x-2}{x-1}\right)-1=\log _{2}\left(\frac{3 x-7}{3 x-1}\right)$, we eliminate the logarithms, simplify the equation, and solve for x. After eliminating the logarithms and simplifying, we obtain the quadratic equation $3x^2 - 7x + 6 = 0$. Factoring the equation gives $(3x-2)(x-3) = 0$, which has two possible solutions, $x = \frac{2}{3}$ and $x = 3$. However, the solution $x = 3$ makes the denominator of the first logarithm equal to 0, so it is not valid. Therefore, the valid solution for the equation is $x = \frac{2}{3}$.

Step by step solution

Eliminate the logarithm

Since both sides of the equation have the same base, we can eliminate the logarithm by setting the arguments of the logarithms equal to each other: \[ \frac{x-2}{x-1} - 1 = \frac{3x-7}{3x-1} \]

Simplify the equation

To simplify the equation, we'll first need to eliminate the fraction: \[ \frac{x-2}{x-1} - \frac{x-1}{x-1} = \frac{3x-7}{3x-1} \] Now, we have: \[ \frac{-1}{x-1} = \frac{3x-7}{3x-1} \] Next, we'll cross-multiply to eliminate the fractions: \[ (-1)(3x-1) = (x-1)(3x-7) \] Expanding both sides: \[ -3x + 1 = 3x^2 - 10x + 7 \]

Solve for x

To solve for x, we'll need to put the equation in the standard form for a quadratic equation: \[ 3x^2 - 7x + 6 = 0 \] Now, we can attempt to factor the equation or use the quadratic formula to find the values of x. In this case, the equation can be factored into: \[ (3x-2)(x-3) = 0 \] So, the equation has two possible solutions: \[ x = \frac{2}{3} \quad\text{or}\quad x = 3 \] However, we must check if these solutions are valid by plugging them back into the original equation and making sure they don't cause the denominator to equal 0. When x=3, the denominator for the first logarithm becomes 0, so we must discard that solution. Therefore, we have only one valid solution for the equation: \[ x = \frac{2}{3} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equation

A quadratic equation is a polynomial equation of the form $ ax^2 + bx + c = 0 $, where $ a $, $ b $, and $ c $ are constants, and $ a eq 0 $. In the context of our original exercise, once we eliminated the logarithms, we were left with a quadratic equation to solve. Quadratic equations are solved for their roots, which are the values of $ x $ that make the equation equal to zero. There are several methods to find these roots:

Factoring: Break down the quadratic equation into a product of two binomials if possible, then set each binomial to zero to find the solutions.
Quadratic Formula: Use the formula $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ to find the roots.
Completing the Square: Transform the quadratic equation into a perfect square trinomial to solve for $ x $.

The original problem utilized factoring to solve the quadratic equation $ 3x^2 - 7x + 6 = 0 $. The factors were found to be $ (3x-2) $ and $ (x-3) $, leading to potential solutions for $ x $.

Logarithm Properties

Logarithms come with several useful properties that can simplify equations, especially when dealing with common bases. In our exercise, the property that $ \log_b(M) - \log_b(N) = \log_b\left(\frac{M}{N}\right) $ was implicitly utilized to initially simplify the given equation. Understanding these properties helps in transforming complex logarithmic statements into more manageable algebraic forms:

Product Property: $ \log_b(MN) = \log_b(M) + \log_b(N) $
Quotient Property: $ \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) $
Power Property: $ \log_b(M^n) = n \cdot \log_b(M) $

For the equation $ \log _{2}\left(\frac{x-2}{x-1}\right)-1=\log _{2}\left(\frac{3x-7}{3x-1}\right) $, recognizing the common base allowed us to equate the contents directly, leading us into a simpler equation to solve.

Solution Verification

Once potential solutions to an equation are found, verifying these solutions is crucial. This step ensures that the solutions are valid and applicable to the original equation. Here's the process of solution verification as demonstrated in our problem:

Substitute Back: Insert the values found for $ x $ back into the original equation to check if they satisfy it.
Check for Extraneous Solutions: Particularly with logarithmic equations, certain values can lead to expressions that are undefined, such as negative arguments or zero denominators. These must be identified and discarded.

In the initial problem, we found two solutions: $ x = \frac{2}{3} $ and $ x = 3 $. However, upon checking, it was clear that $ x = 3 $ made the denominator of the first logarithmic argument zero, leading to an undefined expression. Therefore, the valid solution was only $ x = \frac{2}{3} $. Verification is a vital step to obtaining the correct and realistic solution to mathematical equations.

91影视

$$ \log _{2}\left(\frac{x-2}{x-1}\right)-1=\log _{2}\left(\frac{3 x-7}{3 x-1}\right) $$

Short Answer

Step by step solution

Eliminate the logarithm

Simplify the equation

Solve for x

Key Concepts

Quadratic Equation

Logarithm Properties

Solution Verification

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