Problem 90 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log _{2}(x-1)-\log _{2}(x+3)=\log _{2}\left(\frac{1}{x}\right) $$

Step-by-Step Solution

Answer

The solution to the equation is \(x=3\)

Step 1: Combine the logarithms on the left side

Using properties of logarithms, specifically \(\log_b{a} - \log_b{c} = \log_b{\frac{a}{c}}\), the equation can be transformed into: \(\log_{2}\left(\frac{x-1}{x+3}\right) = \log_{2}\left(\frac{1}{x}\right)\)

Step 2: Equate the arguments of the logarithms

If \(\log_b{a}=\log_b{c}\), then \(a=c\). Equate the arguments of the logarithms to get \(\frac{x-1}{x+3} = \frac{1}{x}\)

Step 3: Solve for \(x\)

Further simplification and solving for \(x\) involves cross-multiplication and gathering terms. Cross multiply to obtain \(x^{2} - x = x + 3\). Gather \(x\) terms on one side and constants on other to obtain: \(x^{2} - 2x - 3 = 0\). This gives us an equation in quadratic form which we can solve by factoring or using the quadratic formula.

Step 4: Factor the quadratic equation

Factoring the quadratic equation \(x^2 - 2x - 3 = 0\), we get \((x-3)(x+1) = 0\). Setting each factor equal to zero gives the solutions \(x=3\) and \(x=-1\)

Step 5: Check the solutions against the original logarithmic expressions

From the solutions \(x=3\) and \(x=-1\), we must discard \(x=-1\) because it doesn't satisfy the condition \(x > 0\) for its corresponding logarithmic expression, \(\log _{2}(x+3)\). Hence, the only solution is \(x=3\)

Step 6: Decimal approximation

Since the exact answer \(x=3\) doesn't require a decimal approximation, we can skip this step for this particular problem.

Key Concepts

Properties of Logarithms

Logarithms have unique properties that simplify complex equations. One vital property is the subtraction of two logarithms, which turns into division. For instance, if you have \(\log_b{a} - \log_b{c}\), you can rewrite it as \(\log_b{\frac{a}{c}}\). This property helps transform logarithmic expressions into easier forms, which is exactly what we used in the exercise.
Some other properties of logarithms include:

• Product Rule: \(\log_b{(a \cdot c)} = \log_b{a} + \log_b{c}\), which helps write logs of products as sums.
• Power Rule: \(\log_b{a^c} = c \cdot \log_b{a}\), allowing us to bring down powers as coefficients.
• Change of Base Formula: \(\log_b{a} = \frac{\log_k{a}}{\log_k{b}}\), useful for converting between different logarithm bases.

Understanding these properties allows us to solve equations, by breaking them into simple components or combining terms to make the equation more manageable.

Domain of Logarithmic Functions

When dealing with logarithmic functions, understanding the domain is crucial. The domain of a logarithmic function is the set of all possible inputs (values of \(x\)) that result in a real number output. For any log function \(\log_b(x)\), the argument \(x\) must be positive (\(x > 0\)).
In our equation, \(\log_2(x-1)\) and \(\log_2(x+3)\) mean that each argument of the logarithm must be greater than zero:

• \(x-1 > 0\) results in \(x > 1\).
• \(x+3 > 0\) results in \(x > -3\).

Thus, the intersection of these conditions is \(x > 1\) for both expressions to be valid.
When solving logarithmic equations, always verify the solutions against the domain to ensure they make sense within the context of the problem. If any solution does not meet the domain criteria, it must be rejected.

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \(ax^2 + bx + c = 0\). Solving these equations is fundamental in mathematics. Quadratics can be solved by various methods:

• Factoring: Breaking down the quadratic into simpler (binomial) expressions which multiply to zoom back to the original equation.
• Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), derived from completing the square of the quadratic.
• Completing the Square: Rewriting the equation in a way that features a perfect square trinomial.

In our problem, the quadratic equation \(x^2 - 2x - 3 = 0\) is factored into \((x - 3)(x + 1) = 0\). Solving gives possible solutions \(x = 3\) and \(x = -1\).
Once factored, set each term equal to zero to find the potential values of \(x\). It is essential to use the domain of the original logarithmic function as a filter to discard any non-viable solutions, as evidenced by rejecting \(x = -1\).

Solution Approximation

Sometimes, solutions need approximation when not precise or when a decimal result is required. In logarithmic equations, solution approximation might involve using a calculator to transform complex roots or expressions into decimals.
However, in the case of the exercise with exact solutions where \(x = 3\), there was no need for decimal approximation. If a solution had required approximation, typically,

• Use a calculator for complex operations.
• Round decimal answers to the specified significant figures, such as two decimal places.
• Ensure consistency in the rounding process to maintain precision.

Understanding when and how to apply approximation techniques ensures that solutions remain accurate and practical to use in further analysis or context.