Problem 51 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{4}(x+3)+\log _{4}(x-3)=2$$

Step-by-Step Solution

Answer

To solve the logarithmic equation \(\log _{4}(x+3)+\log _{4}(x-3)=2\), we first combine the logarithms using the product rule to get \(\log_4{(x+3)(x-3)}=2\). Then, we convert to exponential form and get \(4^2=(x+3)(x-3)\). Solving for x, we find \(x=±5\). However, we must check for extraneous solutions, and we find that \(x=-5\) is not valid. The final solution is \(x=5\), which can be verified using a graphing calculator.

Step 1: Combine the logarithms using the product property

Using the product rule for logarithms, we can combine the two logarithms: \(\log_a{m}+\log_a{n}=\log_a{(mn)}\). In our case, our base is 4 and we have the logarithms of (x+3) and (x-3). So, we get: \(\log _{4}(x+3)+\log _{4}(x-3)=\log_4{(x+3)(x-3)}\) Now, our equation is: \(\log_4{(x+3)(x-3)}=2\)

Step 2: Convert to exponential form

To convert this equation to exponential form, remember that \(\log_a{b}=c\) is equivalent to \(a^c=b\). In this case, our base is 4 and our exponent is 2. So, we get: \(4^2=(x+3)(x-3)\)

Step 3: Solve for x

Now, we can solve for \(x\), by first simplifying the equation: \(16=(x+3)(x-3)\) Next, expanding the equation: \(16=x^2-9\) Then, adding 9 to both sides of the equation: \(x^2=25\) Finally, solving for \(x\) by taking the square root of both sides: \(x=±5\)

Step 4: Check for extraneous solutions

An extraneous solution is a solution that appears to satisfy the given equation but is not actually valid. When dealing with logarithms, we need to make sure that the arguments inside the logarithms are always greater than zero. So, let's check our solutions: For \(x=5\): - \(5+3=8\) - valid - \(5-3=2\) - valid Therefore, \(x=5\) is a valid solution. For \(x=-5\): - \(-5+3=-2\) - invalid - \(-5-3=-8\) - invalid Therefore, \(x=-5\) is not a valid solution.

Step 5: Verify using a graphing calculator

To check our answer, we can graph the equation \(\log_4{(x+3)(x-3)}=2\) using a graphing calculator. When graphing, we should see that the only intersection point between the graph and the horizontal line \(y=2\) will be \(x=5\), confirming that our solution \(x=5\) is correct. Final Answer: \(x=5\)

Key Concepts

Logarithmic Equations

Logarithmic equations involve the use of logarithms, which are the inverse operations of exponentials. Solving these equations often requires manipulating logarithms to isolate the variable of interest. Logarithms have a base, which is the number, under which the other number is operated. In our example, the base is 4. For instance, we can express the equation \( \log_{4}(x+3) + \log_{4}(x-3) = 2 \) by following specific properties of logarithms to eventually make it solvable. Remember, logarithmic expressions are defined only for positive values of their arguments. This aspect becomes crucial when checking if your solutions are valid.

Product Property of Logarithms

The product property of logarithms is a powerful tool for simplifying and solving logarithmic equations. This property states that the logarithm of a product is equal to the sum of the logarithms of the factors:

• \( \log_a{m} + \log_a{n} = \log_a{(mn)} \)

This property allows us to combine multiple logarithmic terms into a single term, making equations easier to handle. In our case, combining \( \log_{4}(x+3) \) and \( \log_{4}(x-3) \) using the product property gives \( \log_4{(x+3)(x-3)} = 2 \). Such simplifications help you transform the logarithmic equation into one that can be more straightforwardly solved, as it frees the equation of multiple logarithmic terms.

Exponential Form

Converting a logarithmic equation into its exponential form is often a crucial step in finding solutions. The general rule for transforming logarithmic to exponential form is:

• \( \log_a{b} = c \) means that \( a^c = b \)

In the example equation, \( \log_4{(x+3)(x-3)} = 2 \), transforming to exponential form gives \( 4^2 = (x+3)(x-3) \). This conversion makes it possible to solve for \( x \) algebraically, by replacing the log expression with a familiar polynomial equation. This is a valuable technique to reduce the complexity inherent in logarithms by leveraging the properties of exponents, making the path to the solution clearer and more intuitive.

Extraneous Solutions

When solving equations, extraneous solutions are solutions that result from the process of solving the equation but do not satisfy the original equation. This happens often when squaring both sides or when using logarithms. In our context, when we solved the equation, we initially got solutions \( x = \pm 5 \). However, not all these solutions are valid when checked in the original logarithmic equation.For logarithmic equations, we must ensure the arguments of the logarithms are positive, because the logarithm of a negative number is undefined. For \( x = 5 \), the arguments \( (x+3) \) and \( (x-3) \) are positive, thus valid.For \( x = -5 \), these arguments become negative, disqualifying \( x = -5 \) as a valid solution. Hence, checking for extraneous solutions is an important step in verifying the correctness of solutions in logarithmic equations.