Problem 44 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log (x+5)-\log (x-3)=\log 2$$

Step-by-Step Solution

Answer

The short answer is as follows: Apply the logarithm property \(\log a - \log b = \log \frac{a}{b}\) to get \(\log \frac{x+5}{x-3} = \log 2\). Next, remove the logarithms by setting the arguments equal to each other: \(\frac{x+5}{x-3} = 2\). Now, solve for x: \(5=x-6\) leading to the solution \(x=11\). Finally, check the solution by substituting x = 11 into the original equation, confirming that both sides equal \(\log 2\).

Step 1: Apply the logarithm property

To simplify the equation, we will use the logarithm property: \(\log a - \log b = \log \frac{a}{b}\). Applying this property, we get: \[\log \frac{x+5}{x-3} = \log 2\]

Step 2: Remove the logarithms

Since both sides of the equation have the same logarithm base, we can remove the logarithms and set the arguments equal to each other: \[\frac{x+5}{x-3} = 2\]

Step 3: Solve for x

To solve for x, follow these steps: 1. Cross-multiply to eliminate the denominator: \[(x+5) = 2(x-3)\] 2. Distribute the constants: \[x+5 = 2x-6\] 3. Move variables to one side of the equation: \[5 = x-6\] 4. Solve for x: \[x = 11\]

Step 4: Check the solution

Now, let's substitute x = 11 into the original equation and check if it satisfies the equation: \[\log(11+5) - \log(11-3) = \log 2\] \[\log 16 - \log 8 = \log 2\] Using properties of logarithms, we have: \[\log \frac{16}{8} = \log 2\] \[\log 2 = \log 2\] Since the equality holds, the solution x = 11 is correct. You can also confirm this by graphing both the original equation in a graphing calculator and seeing that they intersect at x = 11.

Key Concepts

Logarithm Properties

Logarithms can seem tricky, but understanding their basic properties makes solving equations simpler. One property that frequently appears is the difference of logarithms: \( \log a - \log b = \log \frac{a}{b} \). This helps us transform subtraction into division inside the logarithm. This property is a powerful tool for simplifying logarithmic equations.
For example, in our exercise, we start with the equation \( \log (x+5)-\log (x-3)=\log 2 \). Using our logarithm property, we can simplify this to \( \log \frac{x+5}{x-3} = \log 2 \), converting the subtraction of logs into a single expression inside a log.
This not only simplifies the equation but also makes the next steps more straightforward and manageable. Mastering these properties is essential as they frequently provide shortcuts and reduce complexity in logarithmic equations.

Algebraic Solutions

Solving logarithmic equations often involves a few clear algebraic steps. Once we have simplified our logarithmic equation using properties, we can proceed to remove the logarithms. In the equation \( \log \frac{x+5}{x-3} = \log 2 \), as both sides have logarithms with equal bases, we can drop the logs. This gives us \( \frac{x+5}{x-3} = 2 \).
Now, we solve for \( x \) as we would in any algebraic equation. The first step is cross-multiplying to remove the fraction: \((x+5) = 2(x-3)\).
Next, distribute the 2 on the right-hand side: \(x + 5 = 2x - 6\).
Then, gather all terms involving \( x \) to one side, resulting in \( 5 = x - 6 \).
Finally, solve for \( x \) by adding 6 to both sides: \( x = 11 \).
These steps illustrate the blend of logarithmic and algebraic techniques used to find the solution.

Graphing Calculator Verification

Verifying solutions with a graphing calculator is a great way to confirm your work. Once you've solved the logarithmic equation algebraically, it's wise to double-check using technology. To do this, input the left side of the original equation, \( \log(x+5) - \log(x-3) \), and the right side, \( \log 2 \), as separate functions into the calculator.
Then, use the calculator to graph both functions. The intersection point of the graphs represents the solution to the equation. In our case, this occurs at \( x = 11 \).
Seeing where the graphs meet provides visual confirmation of the solution. This step is also useful to catch mistakes you might have made in the algebraic process. It’s a verification step that adds confidence to your results, ensuring everything works as expected.