Problem 75 - 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. $$ \log _{2}(x-6)+\log _{2}(x-4)-\log _{2} x=2 $$

Step-by-Step Solution

Answer

The exact solution to the equation is \(x = 12\).

Step 1: Combine the logarithms

Using the logarithm properties, three logarithms that are added or subtracted can be expressed in terms of division or multiplication in the logarithm. Here, \(\log_{2}(x-6) + \log_{2}(x-4) - \log_{2}(x)\) becomes \(\log_{2}\frac{(x-6)(x-4)}{x}\). Now, the equation is \(\log_{2}\frac{(x-6)(x-4)}{x} = 2\).

Step 2: Convert to an exponential equation

The equation can be rewritten in exponential form: \(2^2 = \frac{(x-6)(x-4)}{x}\), simplifying that gives \(4 = \frac{(x-6)(x-4)}{x}\).

Step 3: Solve the equation

Cross multiplying and simplifying, the equation becomes \(4x = (x-6)(x-4)\). This simplifies further to \(4x = x^2 - 10x + 24\). Making it a quadratic equation, it turns to \(x^2 - 14x + 24 = 0\). Solving for \(x\) using the quadratic formula, gives \(x = 7 \pm 5\). Thus, \(x = 2\) or \(x = 12\). At this point, check any found solution to make sure it lies within the domain of the original logarithmic expressions.

Step 4: Check the solution

The domain of all logs must be greater than zero. Checking for \(x=2\) in the original equation, it can be seen that \(x=2\) is not in the domain of \(\log_{2}(x-6)\) or \(\log_{2}(x-4)\) since \(2-6\) and \(2-4\) are both less than zero. Hence, reject \(x=2\). Checking \(x=12\) now, all the terms \(\log_{2}(12-6)\), \(\log_{2}(12-4)\) and \(\log_{2}(12)\) are valid with \(x=12\) since the arguments are always greater than zero. Therefore, \(x=12\) is the sole solution.

Key Concepts

Logarithm Properties

Understanding logarithm properties is crucial for solving logarithmic equations. Logarithms have several rules that make complex calculations manageable. First, the product rule, which states that the log of a product is equal to the sum of the logs of its factors: \[\begin{equation}\log_b(MN) = \log_b(M) + \log_b(N)\end{equation}\]This property helps in breaking down or combining logarithmic terms. Similarly, the quotient rule allows us to express the logarithm of a division as the difference of logarithms:\[\begin{equation}\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)\end{equation}\]Lastly, the power rule connects exponents within a logarithm to multiplication outside of it:\[\begin{equation}\log_b(M^p) = p\log_b(M)\end{equation}\]Through these properties, complex equations become simpler, as seen in the provided textbook exercise, where the product and quotient rules were used to combine logarithmic terms effectively.

Exponential Form Conversion

At the heart of solving logarithmic equations is the ability to convert between logarithmic and exponential forms. A logarithm equation such as \[\begin{equation}\log_b(M) = N\end{equation}\]can be rewritten as an exponential equation:\[\begin{equation}b^N = M\end{equation}\]This conversion is based on the definition of logarithms, indicating that the base (\[\begin{equation}b\end{equation}\]) raised to the power of the logarithm's result (\[\begin{equation}N\end{equation}\]) equals the argument (\[\begin{equation}M\end{equation}\]). In the exercise, changing the log form to exponential form allowed us to express the equation in terms of multiplication and division, simplifying the subsequent algebra involved in finding the solution.

Quadratic Formula

The quadratic formula is an essential tool for solving equations of the form \[\begin{equation}ax^2 + bx + c = 0\end{equation}\]. It provides a systematic way of finding the roots of any quadratic equation by using the coefficients \[\begin{equation}a\end{equation}\], \[\begin{equation}b\end{equation}\], and \[\begin{equation}c\end{equation}\]:\[\begin{equation}x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{{2a}}\end{equation}\]The discriminant, \[\begin{equation}\sqrt{b^2 - 4ac}\end{equation}\], determines the nature of the roots. If it's positive, there are two real and distinct solutions; if zero, one real solution; and if negative, the solutions are complex. In the exercise, after reducing the logarithmic equation to a quadratic equation, we applied the quadratic formula to find the values of \[\begin{equation}x\end{equation}\].

Domain of Logarithmic Functions

The domain of a logarithmic function consists of all the real numbers for which the function is defined. Since the logarithm of a negative number or zero is undefined, the argument of a logarithmic function must be greater than zero. To find the domain of functions such as \[\begin{equation}\log_b(x-a)\end{equation}\], set the argument greater than zero: \[\begin{equation}(x-a) > 0\end{equation}\], which means \[\begin{equation}x > a\end{equation}\]. Therefore, for the original logarithmic equation, each term needs to have a domain considered. After finding potential solutions by solving the quadratic equation, these solutions are only acceptable if they fall within the domain of the original logarithmic expressions. During the solution check, if any found \[\begin{equation}x-value\end{equation}\] fails this domain test, it must be rejected, as was the case with the value \[\begin{equation}x = 2\end{equation}\] in the exercise.