Problem 56 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{3} x+\log _{3}(x+1)=\log _{3} 2+\log _{3}(x+3)$$

Step-by-Step Solution

Answer

To solve the logarithmic equation \(\log _{3} x+\log _{3}(x+1)=\log _{3} 2+\log _{3}(x+3)\) algebraically, we first apply logarithm properties to simplify the equation: \(\log _{3}(x \cdot (x+1)) = \log _{3}(2 \cdot (x+3))\). Next, remove the logarithm from both sides, expand and reorder the equation to get \(x^2 - x - 6 = 0\). Then, factor the equation \((x-3)(x+2) = 0\) to find the possible solutions, which are \(x=3\) and \(x=-2\). As a logarithm of a negative number is undefined, the valid solution is \(x=3\). We can verify this by graphing the given equations on a graphing calculator, and we will see that the graphs intersect at the point (3, 1.26186), confirming the solution.

Step 1: Use log properties to simplify the equation.

We can apply the product rule for logarithms, which states that: \(\log_a{b} + \log_a{c} = \log_a{(b \cdot c)}\). Therefore, we can rewrite the equation as: \[\log _{3}(x) + \log _{3}(x+1) = \log _{3}(2) + \log _{3}(x+3) \Rightarrow \log _{3}(x \cdot (x+1)) = \log _{3}(2 \cdot (x+3))\]

Step 2: Remove the logarithm from both sides of the equation.

Since both sides of the equation have the same base (\(log _{3}\)), we can remove the logarithm to get: \[x \cdot (x+1) = 2 \cdot (x+3)\]

Step 3: Expand and reorder the equation.

To solve for x, expand the equation and collect all the terms on one side, resulting in a quadratic equation: \[x^2 + x = 2x + 6 \Rightarrow x^2 - x - 6 = 0\]

Step 4: Factor and solve the equation.

Factor the quadratic equation to find the possible solutions(x): \[(x-3)(x+2) = 0\] The possible solutions are: \[x = 3\text{, or } x = -2\]

Step 5: Check which solution is a valid solution.

Since taking a logarithm of a negative number is undefined, x = -2 is not a valid solution. Therefore, the valid solution is: \[x = 3\]

Step 6: Verify with a graphing calculator.

To verify our solution, graph the following equations using a graphing calculator: \[y_1 = \log_3 x + \log_3 (x+1)\] \[y_2 = \log_3 2 + \log_3 (x+3)\] At the intersection point, the coordinates (x, y) should be the same for both graphs. The calculator confirms that these graphs intersect at the point (3, 1.26186), which demonstrates the validity of our algebraically found value of x = 3.

Key Concepts

Product Rule

The product rule for logarithms is a handy tool. It helps you simplify expressions where two logs with the same base are added. The fundamental principle states that if you have

• \( \log_a{b} + \log_a{c} \)

you can rewrite it as:

• \( \log_a{(b \cdot c)} \)

This means you can combine the logs into a single logarithm by multiplying their arguments. This is extremely useful in equations because it reduces complexity.
For the given problem, we used the product rule to combine the terms, resulting in:

• \( \log_3{(x)} + \log_3{(x+1)} = \log_3{(x \cdot (x+1))} \)
• \( \log_3{2} + \log_3{(x+3)} = \log_3{(2 \cdot (x+3))} \)

With the product rule, this process becomes much simpler, allowing for straightforward solutions.

Quadratic Equation

Once the logarithms are removed, you often end up with a quadratic equation. A quadratic equation generally looks like:

• \( ax^2 + bx + c = 0 \)

These can usually be solved by factoring, using the quadratic formula, or by completing the square. In our example, after simplifying the expression, we arrived at:

• \( x^2 - x - 6 = 0 \)

To find the solutions, we factored the quadratic to get:

• \( (x-3)(x+2) = 0 \)

This means the potential solutions are \( x = 3 \) and \( x = -2 \). However, because logarithms of negative numbers are undefined, \( x = -2 \) was not valid. Only \( x = 3 \) works in this context.

Graphing Calculator

A graphing calculator is a powerful tool for verifying algebraic solutions. You can input functions and see their graphs visually, which helps confirm if solutions are correct. In this exercise, we compared the graphs of:

• \( y_1 = \log_3 x + \log_3 (x+1) \)
• \( y_2 = \log_3 2 + \log_3 (x+3) \)

When you plot these, the solution to the equation is found where the graphs intersect. In our case, the intersection point was at \( x = 3 \), confirming what we found algebraically.
Graphing calculators not only help with solving equations but also provide insights into function behaviors that might not be obvious through calculations alone.

Logarithm Properties

Logarithm properties are essential tools when working with logarithmic equations. They include rules like the product rule, quotient rule, and power rule, which simplify expressions and help solve equations. Here's a quick rundown of these properties:

• Product Rule: \( \log_a (bc) = \log_a b + \log_a c \)
• Quotient Rule: \( \log_a \left( \frac{b}{c} \right) = \log_a b - \log_a c \)
• Power Rule: \( \log_a (b^c) = c \cdot \log_a b \)

In our problem, the product rule was primarily used to condense logarithmic expressions. Recognizing and applying these properties is crucial for working efficiently with logs, especially when dealing with both sides of an equation. By transforming the initial equation using the product rule, we moved directly towards solving the problem. These properties are the backbone of simplifying and solving logarithmic expressions.