Problem 90 - Algebra and Trigonometry

Question

Solve each logarithmic equation. 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 \( x \approx 4.54 \).

Step 1: Utilize the properties of logarithms

By using the quotient rule for logarithms, we can combine the left-hand side of the logarithmic equation. The quotient rule states that \(\log_b (\frac{M}{N}) = \log_b M - \log_b N\). Applying this to the given equation, the left-hand side of the equation becomes \(\log_2{ \left(\frac{x-1}{x+3} \right)}\). So, the equation simplifies to: \( \log_2{\left(\frac{x-1}{x+3} \right)} = \log_2{\left(\frac{1}{x} \right)} \).

Step 2: Remove the logarithms

We can now eliminate the logs since the bases are the same. This gives us: \( \frac{x-1}{x+3} = \frac{1}{x}\).

Step 3: Solve the equation

Cross-multiply to solve for x. This results in: \( x^2 - x = 1 + 3x \). Move all terms to one side which gives: \( x^2 - 4x - 1 = 0 \). This is a quadratic equation and can be solved using the quadratic formula \( x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \). Using the quadratic formula gives us two potential solutions for x:

Step 4: Calculate solutions

By using the values a=1, b=-4, and c=-1 in the quadratic formula we get: \( x = \frac{4\pm\sqrt{4^2-4(1)(-1)}}{2*1} \). This simplifies to: \( x\approx4.54, -0.54 \) (to two decimal places).

Step 5: Check for valid solutions

We need to check both solutions to ensure they fall within the domain of the original logarithmic equation. The original equation is defined only when \( x-1 > 0, x+3 > 0, \) and \(x > 0 \). Checking against these restrictions, we find that only x=4.54 is a valid solution, as x=-0.54 does not meet the conditions.

Key Concepts

Properties of Logarithms

The properties of logarithms are essential tools for solving logarithmic equations. Logarithms, being the inverse functions of exponentials, have unique characteristics that can simplify complex expressions and equations.

• Product Rule: The logarithm of a product is equal to the sum of the logarithms of the factors: \[ \log_b(MN) = \log_b(M) + \log_b(N) \].
• Quotient Rule: The logarithm of a quotient is equal to the difference between the logarithm of the numerator and the denominator: \[ \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \].
• Power Rule: The logarithm of a number raised to a power is the power multiplied by the logarithm of the number: \[ \log_b(M^k) = k \cdot \log_b(M) \].
• Change of Base Formula: Allows to convert a logarithm to another base usually to the common base 10 or the natural base e (Euler's number): \[ \log_b(M) = \frac{\log_k(M)}{\log_k(b)} \], where \( k \) is the new base.

In the given exercise, the Quotient Rule was used to combine the logarithmic terms, setting the stage to transform the logarithmic equation into a simpler form. Understanding these properties assists in maneuvering terms effectively, as seen in Step 1 of the solution.

Solving Quadratic Equations

Solving quadratic equations is a cornerstone of algebra. A quadratic equation is typically in the form \[ ax^2 + bx + c = 0 \], where \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero. There are various methods to solve these equations:

• Factoring: Expressing the equation as a product of its factors if they are easily identifiable.
• Square Completing: Rearranging the equation to form a perfect square, thus making it easier to solve.
• Quadratic Formula: A universal method given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], it gives the solutions to any quadratic equation.
• Graphing: Plotting the quadratic function and finding the points where it intersects the x-axis (the roots).

In the solution provided, Step 3 and Step 4 show the quadratic formula in action. It's a reliable technique when factoring is too complex, which can be often the case with non-integer roots or when coefficients are large or unwieldy.

Domain of Logarithmic Functions

The domain of a logarithmic function is the set of all possible values of \( x \) for which the function is defined. For any logarithmic function of the form \[ f(x) = \log_b(x) \], the base \( b \) must be a positive real number not equal to 1, and the argument \( x \) must be positive (\( x > 0 \)).
The domain issues stem from the logarithm's nature; it measures the power to which we must raise the base to obtain the argument. If the argument is zero or negative, it doesn't correspond to any exponent for the positive base, hence the function is undefined.
In the context of the given exercise, it is critical to ensure that the value of \( x \) makes the expressions \( x-1 \) and \( x+3 \) positive to maintain the validity of the logarithmic operations in the original equation. This verification, shown in Step 5, guarantees that the solutions are not only mathematically correct but also applicable within the problem's constraints. Failing to account for the domain can lead to invalid or extraneous solutions.