Select The Correct Answer.What Is The Solution To This Equation?$\ln (x+6) - \ln (2x-1) = 0$A. $x = 7$ B. $x = -7$ C. $x = -5$ D. $x = 5$

Introduction

In this article, we will delve into the world of mathematics and solve a logarithmic equation. The equation in question is $\ln (x+6) - \ln (2x-1) = 0$. We will break down the solution step by step, making it easy to understand and follow along.

Understanding Logarithmic Equations

Before we dive into the solution, let's take a moment to understand what logarithmic equations are. A logarithmic equation is an equation that involves a logarithm, which is the inverse operation of exponentiation. In other words, if $y = \log_b(x)$, then $b^y = x$. Logarithmic equations can be solved using various techniques, including algebraic manipulation and the use of logarithmic properties.

The Given Equation

The given equation is $\ln (x+6) - \ln (2x-1) = 0$. This equation involves two logarithmic terms, which are subtracted from each other. To solve this equation, we will use the properties of logarithms to simplify and manipulate the equation.

Step 1: Simplify the Equation

The first step in solving the equation is to simplify it using the properties of logarithms. Specifically, we will use the property that states $\ln(a) - \ln(b) = \ln(\frac{a}{b})$. Applying this property to the given equation, we get:

$$\ln (x+6) - \ln (2x-1) = \ln \left(\frac{x+6}{2x-1}\right) = 0$$

Step 2: Exponentiate Both Sides

The next step is to exponentiate both sides of the equation. Since the logarithm is the inverse operation of exponentiation, we can "undo" the logarithm by raising both sides to the power of the base of the logarithm, which is $e$ in this case. This gives us:

$$e^{\ln \left(\frac{x+6}{2x-1}\right)} = e^0$$

Simplifying the left-hand side, we get:

$$\frac{x+6}{2x-1} = 1$$

Step 3: Solve for x

The final step is to solve for $x$. To do this, we can start by multiplying both sides of the equation by $2x-1$ to eliminate the fraction. This gives us:

$$x+6 = 2x-1$$

Subtracting $x$ from both sides, we get:

$$6 = x-1$$

Adding $1$ to both sides, we get:

$$7 = x$$

Conclusion

In conclusion, the solution to the equation $\ln (x+6) - \ln (2x-1) = 0$ is $x = 7$. This solution was obtained by simplifying the equation using the properties of logarithms, exponentiating both sides, and solving for $x$.

Answer

The correct answer is:

• A. $x = 7$

Discussion

This equation is a classic example of a logarithmic equation, and it requires a deep understanding of the properties of logarithms to solve. The solution involves simplifying the equation using the properties of logarithms, exponentiating both sides, and solving for $x$. This type of equation is commonly encountered in mathematics and is an important topic to study.

Related Topics

• Logarithmic equations
• Properties of logarithms
• Exponentiation
• Algebraic manipulation

References

• [1] "Logarithmic Equations" by Math Open Reference
• [2] "Properties of Logarithms" by Khan Academy
• [3] "Exponentiation" by Wolfram MathWorld
• [4] "Algebraic Manipulation" by MIT OpenCourseWare
  Frequently Asked Questions: Logarithmic Equations =====================================================

Introduction

In our previous article, we solved the equation $\ln (x+6) - \ln (2x-1) = 0$. In this article, we will answer some frequently asked questions related to logarithmic equations. These questions cover various topics, including the properties of logarithms, exponentiation, and algebraic manipulation.

Q: What is the difference between a logarithmic equation and an exponential equation?

A: A logarithmic equation is an equation that involves a logarithm, which is the inverse operation of exponentiation. An exponential equation, on the other hand, is an equation that involves an exponent. For example, the equation $\ln (x+6) - \ln (2x-1) = 0$ is a logarithmic equation, while the equation $2^x = 8$ is an exponential equation.

Q: How do I simplify a logarithmic equation?

A: To simplify a logarithmic equation, you can use the properties of logarithms, such as the product rule, quotient rule, and power rule. For example, the equation $\ln (x+6) - \ln (2x-1) = 0$ can be simplified using the quotient rule, which states that $\ln(a) - \ln(b) = \ln(\frac{a}{b})$.

Q: How do I exponentiate both sides of a logarithmic equation?

A: To exponentiate both sides of a logarithmic equation, you can raise both sides to the power of the base of the logarithm. For example, if the equation is $\ln (x+6) - \ln (2x-1) = 0$, you can exponentiate both sides by raising both sides to the power of $e$, which is the base of the natural logarithm.

Q: How do I solve for x in a logarithmic equation?

A: To solve for x in a logarithmic equation, you can use algebraic manipulation to isolate x. For example, if the equation is $\ln (x+6) - \ln (2x-1) = 0$, you can simplify the equation using the properties of logarithms, exponentiate both sides, and then solve for x.

Q: What are some common mistakes to avoid when solving logarithmic equations?

A: Some common mistakes to avoid when solving logarithmic equations include:

• Not using the properties of logarithms correctly
• Not exponentiating both sides of the equation
• Not isolating x correctly
• Not checking the domain of the logarithmic function

Q: How do I check the domain of a logarithmic function?

A: To check the domain of a logarithmic function, you need to make sure that the argument of the logarithm is positive. For example, if the equation is $\ln (x+6) - \ln (2x-1) = 0$, you need to make sure that $x+6 > 0$ and $2x-1 > 0$.

Conclusion

In conclusion, logarithmic equations are an important topic in mathematics, and solving them requires a deep understanding of the properties of logarithms, exponentiation, and algebraic manipulation. By following the steps outlined in this article, you can solve logarithmic equations and avoid common mistakes.

Related Topics

• Logarithmic equations
• Properties of logarithms
• Exponentiation
• Algebraic manipulation
• Domain of a logarithmic function

References

• [1] "Logarithmic Equations" by Math Open Reference
• [2] "Properties of Logarithms" by Khan Academy
• [3] "Exponentiation" by Wolfram MathWorld
• [4] "Algebraic Manipulation" by MIT OpenCourseWare