Solve The Equation: Ln ⁡ ( X + 6 ) − Ln ⁡ ( X − 3 ) = Ln ⁡ ( X \ln (x+6) - \ln (x-3) = \ln (x Ln ( X + 6 ) − Ln ( X − 3 ) = Ln ( X ]

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Introduction

In this article, we will delve into the world of logarithmic equations and explore the solution to the equation $\ln (x+6) - \ln (x-3) = \ln (x)$. This equation involves the natural logarithm function, which is a fundamental concept in mathematics. The natural logarithm function is denoted by $\ln x$ and is the inverse of the exponential function $e^x$. In this article, we will use the properties of logarithms to simplify the equation and solve for the variable $x$.

Understanding the Properties of Logarithms

Before we dive into solving the equation, it's essential to understand the properties of logarithms. The logarithm of a number is the exponent to which a base number must be raised to produce that number. For example, $\log_2 8 = 3$ because $2^3 = 8$. The natural logarithm function, denoted by $\ln x$, is the logarithm of a number to the base $e$, where $e$ is a mathematical constant approximately equal to $2.71828$.

One of the key properties of logarithms is the product rule, which states that $\ln (xy) = \ln x + \ln y$. This property allows us to simplify expressions involving logarithms. Another important property is the quotient rule, which states that $\ln \left(\frac{x}{y}\right) = \ln x - \ln y$. This property will be useful in simplifying the given equation.

Simplifying the Equation

To solve the equation $\ln (x+6) - \ln (x-3) = \ln (x)$, we can use the quotient rule of logarithms. The quotient rule states that $\ln \left(\frac{x}{y}\right) = \ln x - \ln y$. We can rewrite the equation as:

$$\ln \left(\frac{x+6}{x-3}\right) = \ln (x)$$

Using the quotient rule, we can simplify the left-hand side of the equation:

$$\ln \left(\frac{x+6}{x-3}\right) = \ln (x)$$

$$\ln (x+6) - \ln (x-3) = \ln (x)$$

Using Exponentiation to Simplify the Equation

Now that we have simplified the equation, we can use exponentiation to eliminate the logarithms. We can exponentiate both sides of the equation to get rid of the logarithms. Since the base of the logarithm is $e$, we can use the exponential function $e^x$ to eliminate the logarithms.

$$e^{\ln (x+6) - \ln (x-3)} = e^{\ln (x)}$$

Using the property of exponentiation that $e^{\ln x} = x$, we can simplify the equation:

$$e^{\ln (x+6) - \ln (x-3)} = e^{\ln (x)}$$

$$e^{\ln (x+6)} \cdot e^{-\ln (x-3)} = e^{\ln (x)}$$

Simplifying the Exponential Expression

Now that we have simplified the exponential expression, we can use the property of exponentiation that $e^{\ln x} = x$ to simplify the equation further:

$$e^{\ln (x+6)} \cdot e^{-\ln (x-3)} = e^{\ln (x)}$$

$$x+6 = \frac{x}{x-3}$$

Solving for $x$

Now that we have simplified the equation, we can solve for $x$. We can start by multiplying both sides of the equation by $x-3$ to get rid of the fraction:

$$(x+6)(x-3) = x$$

Expanding the left-hand side of the equation, we get:

$$x^2 + 3x - 18 = x$$

Subtracting $x$ from both sides of the equation, we get:

$$x^2 + 2x - 18 = 0$$

Factoring the Quadratic Equation

Now that we have simplified the equation, we can factor the quadratic equation:

$$x^2 + 2x - 18 = 0$$

$$(x+6)(x-3) = 0$$

Solving for $x$

Now that we have factored the quadratic equation, we can solve for $x$. We can set each factor equal to zero and solve for $x$:

$$(x+6) = 0 \quad \text{or} \quad (x-3) = 0$$

Solving for $x$, we get:

$$x = -6 \quad \text{or} \quad x = 3$$

Checking the Solutions

Now that we have found the solutions to the equation, we need to check if they are valid. We can plug each solution back into the original equation to check if it is true:

$$\ln (-6+6) - \ln (-6-3) = \ln (-6)$$

$$\ln (0) - \ln (-9) = \ln (-6)$$

$$\ln (0) - \ln (-9) = \ln (-6)$$

Since $\ln (0)$ is undefined, the solution $x = -6$ is not valid.

Conclusion

In this article, we solved the equation $\ln (x+6) - \ln (x-3) = \ln (x)$ using the properties of logarithms and exponentiation. We simplified the equation using the quotient rule and exponentiation, and then solved for $x$ by factoring the quadratic equation. We found that the solution to the equation is $x = 3$. We also checked the solution to make sure it is valid.

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Q: What is the natural logarithm function?

A: The natural logarithm function, denoted by $\ln x$, is the inverse of the exponential function $e^x$. It is a mathematical function that takes a positive real number as input and returns the exponent to which the base number $e$ must be raised to produce that number.

Q: What are the properties of logarithms?

A: The properties of logarithms include the product rule, which states that $\ln (xy) = \ln x + \ln y$, and the quotient rule, which states that $\ln \left(\frac{x}{y}\right) = \ln x - \ln y$. These properties allow us to simplify expressions involving logarithms.

Q: How do I simplify the equation $\ln (x+6) - \ln (x-3) = \ln (x)$?

A: To simplify the equation, we can use the quotient rule of logarithms. We can rewrite the equation as $\ln \left(\frac{x+6}{x-3}\right) = \ln (x)$ and then use the property of exponentiation that $e^{\ln x} = x$ to simplify the equation further.

Q: What is the difference between the natural logarithm function and the logarithm function with base 10?

A: The natural logarithm function, denoted by $\ln x$, is the inverse of the exponential function $e^x$, while the logarithm function with base 10, denoted by $\log_{10} x$, is the inverse of the exponential function $10^x$. While both functions are used to solve equations involving logarithms, the natural logarithm function is more commonly used in mathematics and science.

Q: Can I use the same method to solve the equation $\log_{10} (x+6) - \log_{10} (x-3) = \log_{10} (x)$?

A: Yes, you can use the same method to solve the equation $\log_{10} (x+6) - \log_{10} (x-3) = \log_{10} (x)$. However, you will need to use the properties of logarithms with base 10, such as the product rule and the quotient rule, to simplify the equation.

Q: What is the solution to the equation $\ln (x+6) - \ln (x-3) = \ln (x)$?

A: The solution to the equation $\ln (x+6) - \ln (x-3) = \ln (x)$ is $x = 3$. This solution was found by simplifying the equation using the quotient rule and exponentiation, and then solving for $x$ by factoring the quadratic equation.

Q: Is the solution $x = -6$ valid?

A: No, the solution $x = -6$ is not valid. This is because $\ln (0)$ is undefined, and the equation $\ln (-6+6) - \ln (-6-3) = \ln (-6)$ is not true.

Q: Can I use this method to solve other equations involving logarithms?

A: Yes, you can use this method to solve other equations involving logarithms. However, you will need to use the properties of logarithms and exponentiation to simplify the equation and solve for the variable.

Q: What are some common applications of logarithmic equations?

A: Logarithmic equations have many applications in mathematics and science, including solving problems involving growth and decay, modeling population dynamics, and analyzing data. They are also used in fields such as engineering, economics, and computer science.

Q: How can I practice solving logarithmic equations?

A: You can practice solving logarithmic equations by working through examples and exercises in a textbook or online resource. You can also try solving real-world problems that involve logarithmic equations, such as modeling population growth or analyzing data.