Problem 48 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{3}(x+14)-\log _{3}(x+6)=\log _{3} x$$

Step-by-Step Solution

Answer

To solve the logarithmic equation algebraically, we first combined the logs: \(\log_3{\left (\frac{x+14}{x+6}\right )} = \log_3 x\). Then, we removed the logs, obtaining the equation: \(\frac{x+14}{x+6} = x\). We solved this equation for x, finding two potential solutions: x=2 and x=-7. However, x=-7 would result in the log of a negative number, so we only consider the solution x=2. We can verify our solution using a graphing calculator by graphing \(y_1 = \log_3{(x+14)} - \log_3{(x+6)}\) and \(y_2 = \log_3{x}\), and observing the intersection point with an x-coordinate of 2, confirming our solution is correct.

Step 1: Combine the Logs

To combine the logs, we will use the logarithmic property: \(\log_a{M} - \log_a{N} = \log_a{\left (\frac{M}{N} \right )}\). Applying this property in our equation: \[\log_3{\left ( \frac{x+14}{x+6} \right)} = \log_3 x\]

Step 2: Remove Logs

Since the base and the logarithm are identical on both sides, we can take the \(\log_3\) out: \[\frac{x+14}{x+6} = x\]

Step 3: Solve for x

Now we can solve for x. Firstly, we will eliminate the denominator by cross-multiplying: \[x+14 = x(x+6)\] Now, expand the right side of the equation: \[x+14 = x^2 + 6x\] Move all terms to one side to create a quadratic equation: \[x^2 + 5x - 14 = 0\] Factor this quadratic equation: \[(x-2)(x+7) = 0\] Now, find the possible solutions: \(x-2 = 0\) or \(x+7 = 0\) \(x = 2\) or \(x = -7\)

Step 4: Check the Solutions

Looking back at the original equation, we see that a value of x=-7 would result in the log of a negative number, which is undefined. Therefore, we disregard this solution, and our final solution is x=2.

Step 5: Verify with a Graphing Calculator

To verify our solution, we will use a graphing calculator to graph the two functions: \(y_1 = \log_3{(x+14)}-\log_3{(x+6)}\) and \(y_2 = \log_3{x}\). Now, look for the point of intersection between the graphs. The x-coordinate of the intersection point should be the solution of our original equation. If the graphing calculator indicates that the intersection point has an x-coordinate of 2, then our solution (x=2) is correct.

Key Concepts

Logarithmic Properties

Understanding the logarithmic properties is fundamental when solving logarithmic equations. One key property we use frequently in these problems is the quotient rule. The quotient rule states that for any positive numbers \(M\) and \(N\), and a base \(a\), \ \(\log_a{M} - \log_a{N} = \log_a{\left(\frac{M}{N}\right)}\). This allows us to condense the difference of two logarithms into a single logarithm expression.

• This property is useful because it simplifies our equations, reducing the number of terms and making it easier to solve the equation.
• This property emphasizes that subtraction of logs translates into division of the arguments.

Let's apply this property to the logarithmic equation \(\log_3(x+14) - \log_3(x+6) = \log_3 x\).
By using the quotient rule, we simplify the left side to \(\log_3\left(\frac{x+14}{x+6}\right)\), helping us move a step closer to solving the equation algebraically.

Solving Equations Algebraically

To solve the logarithmic equation algebraically, merging terms using properties is just the first step. In the given equation, once we have \(\log_3\left(\frac{x+14}{x+6}\right) = \log_3 x\), we can "drop" the logarithms, due to their identical bases.

• Dropping the logs is permissible because if \(\log_a M = \log_a N\), then \(M = N\).
• Doing so reduces the complexity considerably as we now have a simple rational equation \(\frac{x+14}{x+6} = x\).

The next step involves solving the rational equation. Cross-multiplication helps eliminate the fraction:
\(x + 14 = x(x + 6)\). It leads to a quadratic equation \(x^2 + 6x = x + 14\) when expanded.

Quadratic Equations

Quadratic equations play a crucial role in this kind of logarithmic problem where rational expressions are involved. Once you convert the logarithmic equation into the quadratic form \(x^2 + 5x - 14 = 0\), you can solve it using algebraic methods such as factoring, the quadratic formula, or completing the square.

• In our case, factoring is convenient: \((x - 2)(x + 7) = 0\).
• This provides potential solutions \(x = 2\) and \(x = -7\).

It is important to verify these solutions in the context of the original problem because solutions where substitutions result in taking a log of a negative or zero are invalid. Here, \(x = -7\) yields invalid logs and must be discarded, leaving \(x = 2\) as the solution.

Graphing Calculator Verification

Verifying algebraic solutions with a graphing calculator ensures accuracy and provides a visual confirmation of the work completed. For the original equation \(\log_3(x+14) - \log_3(x+6) = \log_3 x\), plot the functions \(y_1 = \log_3(x+14) - \log_3(x+6)\) and \(y_2 = \log_3 x\) on the calculator.

• Graphing both functions allows you to visually locate their intersection point.
• The \(x\)-coordinate of this intersection should coincide with the solution of the equation.

In our case, if the graphs intersect at \(x = 2\), the solution \(x = 2\) is confirmed to be correct. This process reassures you of the accuracy of the solution found through algebraic means.