Problem 18 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$ \log _{4}(3 x+2)=\log _{4} 5+\log _{4} 3 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{13}{3} \).

Step 1: Apply the Property of Logarithms

Start by recognizing the property of logarithms that allows us to combine the terms on the right side of the equation. The product property states \( \log_b M + \log_b N = \log_b (MN) \). Thus, rewrite the right-hand side as \( \log_{4}(5 \times 3) \).

Step 2: Simplify the Right-Hand Side

Calculate the product within the logarithmic expression on the right: \( 5 \times 3 = 15 \). So, rewrite the right side as \( \log_{4} 15 \). The equation now is \( \log _{4}(3x+2) = \log _{4} 15 \).

Step 3: Equate the Arguments of the Logarithms

Since the logarithms are of the same base and their arguments are equal, we can equate the expressions inside the logarithms: \( 3x + 2 = 15 \).

Step 4: Solve for x

Subtract 2 from both sides to isolate the term with \( x \): \( 3x = 13 \). Then, divide both sides by 3 to solve for \( x \): \( x = \frac{13}{3} \).

Step 5: Verify the Solution

Substitute \( x = \frac{13}{3} \) back into the original equation to ensure both sides equal when simplified. Verification shows that the equation holds true.

Key Concepts

Properties of Logarithms

Logarithms play an essential role in solving exponential equations and transforming multiplication into addition, which is often more manageable. One crucial property is the **Product Property**, expressed as:

• \( \log_b M + \log_b N = \log_b (MN) \)

This property allows us to combine two logarithmic expressions into one by multiplying their arguments, as long as they share the same base.
In the original exercise, we encountered two separate logarithmic terms with the same base, \( \log_4 5 \) and \( \log_4 3 \). By applying the Product Property, we combined these into a single logarithmic term:

• \( \log_4 5 + \log_4 3 = \log_4 (5 \times 3) = \log_4 15 \)

This property simplifies the equation significantly, reducing complexity and making it easier to solve. Understanding how to apply these properties is key in tackling logarithmic equations.

Equating Logarithmic Expressions

Once we simplify an equation using logarithmic properties, the next step often involves equating the logarithmic expressions. This process works smoothly when the logarithms have the same base. In the given problem, after simplification, the equation becomes:

• \( \log_4(3x + 2) = \log_4 15 \)

With their bases now identical and balanced, we can now focus on the arguments. Since the logarithms of the same base must have equal arguments to be true, we equate the expression inside the logarithms:

• \( 3x + 2 = 15 \)

This equating method is very useful. Solving for the variable in logarithmic equations often relies on equating the arguments like this once simplification is done.

Solving for Variables

Solving for variables in logarithmic equations often involves simple algebraic manipulations after equating the arguments. Here, once the equation \( 3x + 2 = 15 \) is formed, the next step is to isolate the variable \( x \).

• First, subtract 2 from both sides: \( 3x = 13 \)
• Then, divide both sides by 3: \( x = \frac{13}{3} \)

These straightforward steps are typical in solving linear equations, often encountered when solving logarithmic equations. Finally, it is important to verify solutions by substituting back into the original equation to confirm that both sides of the equation hold true, ensuring the solution is correct. Hence, consistent practice of solving and verifying ensures mastery of understanding logarithmic equations.