Problem 340 - Precalculus

Question

Solve each equation for \(x.\) \(\log _{3}(3 x)-\log _{3}(6)=\log _{3}(77)\)

Step-by-Step Solution

Answer

The solution for \( x \) is 154.

Step 1: Apply Logarithm Properties

First, let's use the property of logarithms that states \( \log_b(M) - \log_b(N) = \log_b\left(\frac{M}{N}\right) \). Applying this to the equation \( \log_3(3x) - \log_3(6) \), we can rewrite it as: \[ \log_3\left(\frac{3x}{6}\right) = \log_3(77) \] This simplifies to \( \log_3\left(\frac{x}{2}\right) = \log_3(77) \).

Step 2: Equate the Arguments

Since we have \( \log_3\left(\frac{x}{2}\right) = \log_3(77) \), the arguments of the logarithms must be equal. Therefore, we set: \[ \frac{x}{2} = 77 \] This step involves equating the inside of the logarithm expressions because if \( \log_b(A) = \log_b(B) \), then \( A = B \).

Step 3: Solve for x

To solve \( \frac{x}{2} = 77 \), multiply both sides of the equation by 2 to isolate \( x \): \[ x = 77 \times 2 \] \[ x = 154 \] This gives the value of \( x \) that satisfies the original equation.

Key Concepts

Logarithm Properties

Understanding logarithm properties is key to solving a variety of equations involving logarithms. One of the most commonly used properties is the difference of logarithms, which states:

• \( \log_b(M) - \log_b(N) = \log_b\left(\frac{M}{N}\right) \)

This property allows us to combine two separate logarithm terms into a single term. In the given exercise, this property was used to rewrite \( \log_3(3x) - \log_3(6) \) as \( \log_3\left(\frac{3x}{6}\right) \). This simplifies the equation significantly.
Applying these properties helps in reducing complexity and makes it easier to solve equations by eliminating one of the variables early on in the process.

Solving Equations

Solving equations with logarithms follow similar principles as solving any algebraic equations, but incorporate understanding of logarithm properties. Once you've simplified the equation using logarithm properties, as in the step from \( \log_3\left(\frac{x}{2}\right) = \log_3(77) \), the next step is determining the values for which both sides of the equation are equal.

Equating the Arguments

When the bases and logarithmic values are equal, we can equate their arguments:

• If \( \log_b(A) = \log_b(B) \), then \( A = B \).

This simple rule makes solving logarithmic equations more straightforward. In the problem, equating \( \frac{x}{2} \) and 77 allows us to solve for \( x \) using elementary algebraic methods.

Change of Base Formula

The change of base formula is a useful tool when you want to evaluate logarithms with different bases, especially when a calculator only supports common log bases such as 10 or \( e \). The formula is expressed as:

• \( \log_b(A) = \frac{\log_c(A)}{\log_c(b)} \)

This formula allows you to convert a logarithm to a different base, making calculations easier and allowing more flexibility in solving equations. Although it wasn't directly used in the specific problem, understanding this concept enhances your problem-solving toolkit and allows tackling more complex logarithmic equations in precalculus.

Precalculus

Precalculus serves as a bridge between algebra and calculus, covering a variety of topics, including logarithms. Logarithmic equations are a fundamental part of this subject. Proficiency in these topics prepares students for advanced mathematical concepts and applications in calculus.

Applying Logarithms in Precalculus

In precalculus, students learn to manipulate and solve equations involving logarithms and exponential functions. This solidifies their algebra skills while introducing them to the broader world of mathematical analysis. Concepts like the logarithm properties and change of base formula are not just academic exercises; they are pivotal in understanding growth models, financial calculations, and even in fields like physics and engineering.
Grasping logarithmic concepts deeply at this stage is crucial for success in subsequent mathematical studies.