Solve For $x$: Log ⁡ ( X + 4 ) − Log ⁡ ( X + 2 ) = 1 \log (x+4)-\log (x+2)=1 Lo G ( X + 4 ) − Lo G ( X + 2 ) = 1 X = X= X = You May Enter The Exact Value Or Round To 4 Decimal Places.

Introduction

Logarithmic equations can be challenging to solve, but with the right approach, they can be broken down into manageable steps. In this article, we will focus on solving the equation $\log (x+4)-\log (x+2)=1$. This equation involves logarithms with different bases, and our goal is to isolate the variable x and find its value.

Understanding Logarithmic Properties

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 $2^3 = 8$. Logarithmic properties are crucial in solving logarithmic equations, and we will use them extensively in this article.

The Difference of Logarithms

The equation $\log (x+4)-\log (x+2)=1$ involves the difference of logarithms. According to the logarithmic properties, the difference of logarithms can be rewritten as a single logarithm. Specifically, $\log a - \log b = \log \frac{a}{b}$. Applying this property to our equation, we get:

$$\log \frac{x+4}{x+2} = 1$$

Exponentiating Both Sides

To eliminate the logarithm, we can exponentiate both sides of the equation. Since the base of the logarithm is not specified, we will assume that it is the natural logarithm (base e). Exponentiating both sides gives us:

$$e^{\log \frac{x+4}{x+2}} = e^1$$

Simplifying the Equation

Using the property of logarithms that $e^{\log a} = a$, we can simplify the left-hand side of the equation:

$$\frac{x+4}{x+2} = e^1$$

Solving for x

Now we have a simple algebraic equation to solve. We can start by multiplying both sides by $x+2$ to eliminate the fraction:

$$x+4 = e^1(x+2)$$

Expanding and Simplifying

Expanding the right-hand side of the equation gives us:

$$x+4 = ex + 2e$$

Isolating x

Subtracting $x$ from both sides gives us:

$$4 = ex + 2e - x$$

Rearranging Terms

Rearranging the terms gives us:

$$4 = x(e-1) + 2e$$

Factoring Out x

Factoring out x gives us:

$$4 = x(e-1) + 2e$$

Solving for x

Now we can solve for x by isolating it on one side of the equation. Subtracting $2e$ from both sides gives us:

$$4 - 2e = x(e-1)$$

Dividing Both Sides

Dividing both sides by $(e-1)$ gives us:

$$\frac{4 - 2e}{e-1} = x$$

Simplifying the Expression

Simplifying the expression gives us:

$$\frac{4 - 2e}{e-1} = x$$

Evaluating the Expression

Evaluating the expression gives us:

$$x = \frac{4 - 2e}{e-1}$$

Rounding to 4 Decimal Places

Rounding the value of x to 4 decimal places gives us:

$$x = 1.3097$$

Conclusion

In this article, we solved the logarithmic equation $\log (x+4)-\log (x+2)=1$ using logarithmic properties and algebraic manipulations. We started by rewriting the equation as a single logarithm, then exponentiated both sides to eliminate the logarithm. We simplified the equation and isolated x to find its value. The final value of x is $1.3097$, rounded to 4 decimal places.

Final Answer

The final answer is: $\boxed{1.3097}$

Introduction

In our previous article, we solved the logarithmic equation $\log (x+4)-\log (x+2)=1$ using logarithmic properties and algebraic manipulations. In this article, we will provide a Q&A guide to help you understand the solution and answer any questions you may have.

Q: What is the difference of logarithms?

A: The difference of logarithms is a logarithmic property that states $\log a - \log b = \log \frac{a}{b}$. This property allows us to rewrite the difference of logarithms as a single logarithm.

Q: How do I apply the difference of logarithms property to the equation?

A: To apply the difference of logarithms property, you need to rewrite the equation as a single logarithm. In this case, we have $\log (x+4)-\log (x+2)=1$. We can rewrite this as $\log \frac{x+4}{x+2} = 1$.

Q: What is the next step in solving the equation?

A: The next step is to exponentiate both sides of the equation. Since the base of the logarithm is not specified, we will assume that it is the natural logarithm (base e). Exponentiating both sides gives us $e^{\log \frac{x+4}{x+2}} = e^1$.

Q: How do I simplify the equation after exponentiating both sides?

A: After exponentiating both sides, we can simplify the equation using the property of logarithms that $e^{\log a} = a$. This gives us $\frac{x+4}{x+2} = e^1$.

Q: What is the value of $e^1$?

A: The value of $e^1$ is approximately 2.71828.

Q: How do I solve for x?

A: To solve for x, we can start by multiplying both sides of the equation by $x+2$ to eliminate the fraction. This gives us $x+4 = e^1(x+2)$.

Q: How do I simplify the equation further?

A: We can simplify the equation further by expanding the right-hand side and rearranging the terms. This gives us $x+4 = ex + 2e$.

Q: How do I isolate x?

A: To isolate x, we can subtract $x$ from both sides of the equation and rearrange the terms. This gives us $4 = x(e-1) + 2e$.

Q: How do I solve for x?

A: To solve for x, we can divide both sides of the equation by $(e-1)$ and simplify the expression. This gives us $x = \frac{4 - 2e}{e-1}$.

Q: What is the final value of x?

A: The final value of x is approximately 1.3097, rounded to 4 decimal places.

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

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

• Not applying the difference of logarithms property correctly
• Not exponentiating both sides of the equation
• Not simplifying the equation correctly
• Not isolating x correctly

Q: How can I practice solving logarithmic equations?

A: You can practice solving logarithmic equations by working through example problems and exercises. You can also try solving logarithmic equations with different bases and exponents.

Conclusion

In this article, we provided a Q&A guide to help you understand the solution to the logarithmic equation $\log (x+4)-\log (x+2)=1$. We covered common mistakes to avoid and provided tips for practicing solving logarithmic equations. We hope this guide has been helpful in understanding the solution and answering any questions you may have.