Which Of The Following Shows The Extraneous Solution To The Logarithmic Equation Below?${ 2 \log _5(x+1) = 2 }$A. { X = -6 $}$B. { X = -4 $}$C. { X = -2 $}$D. { X = -1 $}$

Introduction

Logarithmic equations are a fundamental concept in mathematics, and solving them requires a deep understanding of the properties of logarithms. In this article, we will focus on solving a specific logarithmic equation and identifying the extraneous solution. We will use the given equation: $2 \log_5(x+1) = 2$ and determine which of the provided options is the extraneous solution.

Understanding Logarithmic Equations

A logarithmic equation is an equation that involves a logarithm. The general form of a logarithmic equation is $\log_b(x) = y$, where $b$ is the base of the logarithm, $x$ is the argument, and $y$ is the result. In this equation, the logarithm is the inverse operation of exponentiation.

Solving the Given Equation

To solve the given equation, we need to isolate the variable $x$. We can start by using the property of logarithms that states $\log_b(x) = y \iff b^y = x$. In this case, we have:

$$2 \log_5(x+1) = 2$$

Using the property of logarithms, we can rewrite the equation as:

$$5^2 = x+1$$

Simplifying the equation, we get:

$$25 = x+1$$

Subtracting 1 from both sides, we get:

$$24 = x$$

However, this is not the only solution to the equation. We need to check if there are any extraneous solutions.

Identifying Extraneous Solutions

An extraneous solution is a solution that is not valid for the original equation. In this case, we need to check if the solution $x = 24$ is valid for the original equation.

Substituting $x = 24$ into the original equation, we get:

$$2 \log_5(24+1) = 2$$

Simplifying the equation, we get:

$$2 \log_5(25) = 2$$

Using the property of logarithms, we can rewrite the equation as:

$$\log_5(25^2) = 2$$

Simplifying the equation, we get:

$$\log_5(625) = 2$$

This is a true statement, so $x = 24$ is a valid solution.

However, we need to check the other options to see if they are extraneous solutions.

Checking the Other Options

Let's check the other options to see if they are extraneous solutions.

Option A: $x = -6$

Substituting $x = -6$ into the original equation, we get:

$$2 \log_5(-6+1) = 2$$

Simplifying the equation, we get:

$$2 \log_5(-5) = 2$$

This is a false statement, so $x = -6$ is an extraneous solution.

Option B: $x = -4$

Substituting $x = -4$ into the original equation, we get:

$$2 \log_5(-4+1) = 2$$

Simplifying the equation, we get:

$$2 \log_5(-3) = 2$$

This is a false statement, so $x = -4$ is an extraneous solution.

Option C: $x = -2$

Substituting $x = -2$ into the original equation, we get:

$$2 \log_5(-2+1) = 2$$

Simplifying the equation, we get:

$$2 \log_5(-1) = 2$$

This is a false statement, so $x = -2$ is an extraneous solution.

Option D: $x = -1$

Substituting $x = -1$ into the original equation, we get:

$$2 \log_5(-1+1) = 2$$

Simplifying the equation, we get:

$$2 \log_5(0) = 2$$

This is a false statement, so $x = -1$ is an extraneous solution.

Conclusion

In this article, we solved a logarithmic equation and identified the extraneous solution. We used the property of logarithms to rewrite the equation and isolate the variable $x$. We then checked the other options to see if they were extraneous solutions. We found that $x = -6$, $x = -4$, $x = -2$, and $x = -1$ were all extraneous solutions.

Final Answer

The final answer is:

• A. $x = -6$

Introduction

Logarithmic equations are a fundamental concept in mathematics, and solving them requires a deep understanding of the properties of logarithms. In this article, we will provide a Q&A guide to help you understand logarithmic equations and how to solve them.

Q: What is a logarithmic equation?

A: A logarithmic equation is an equation that involves a logarithm. The general form of a logarithmic equation is $\log_b(x) = y$, where $b$ is the base of the logarithm, $x$ is the argument, and $y$ is the result.

Q: How do I solve a logarithmic equation?

A: To solve a logarithmic equation, you need to isolate the variable $x$. You can use the property of logarithms that states $\log_b(x) = y \iff b^y = x$. This means that you can rewrite the logarithmic equation as an exponential equation and solve for $x$.

Q: What is an extraneous solution?

A: An extraneous solution is a solution that is not valid for the original equation. In other words, it is a solution that does not satisfy the original equation.

Q: How do I identify an extraneous solution?

A: To identify an extraneous solution, you need to substitute the solution into the original equation and check if it is true. If the solution does not satisfy the original equation, then it is an extraneous solution.

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 correct property of logarithms
• Not isolating the variable $x$
• Not checking for extraneous solutions
• Not using the correct base for the logarithm

Q: How do I choose the correct base for a logarithm?

A: The base of a logarithm is the number that is used to raise the argument to a power. For example, in the equation $\log_5(x) = y$, the base is 5. You can choose any base for a logarithm, but it is usually best to choose a base that is easy to work with.

Q: What are some common logarithmic equations?

A: Some common logarithmic equations include:

• $\log_b(x) = y$
• $\log_b(x) = \log_b(z)$
• $\log_b(x) = \log_b(z) + \log_b(w)$

Q: How do I solve a logarithmic equation with multiple logarithms?

A: To solve a logarithmic equation with multiple logarithms, you need to use the property of logarithms that states $\log_b(x) + \log_b(y) = \log_b(xy)$. This means that you can combine the multiple logarithms into a single logarithm and solve for $x$.

Q: What are some real-world applications of logarithmic equations?

A: Logarithmic equations have many real-world applications, including:

• Finance: Logarithmic equations are used to calculate interest rates and investment returns.
• Science: Logarithmic equations are used to model population growth and decay.
• Engineering: Logarithmic equations are used to design and optimize systems.

Conclusion

In this article, we provided a Q&A guide to help you understand logarithmic equations and how to solve them. We covered topics such as the definition of a logarithmic equation, how to solve a logarithmic equation, and how to identify an extraneous solution. We also discussed common mistakes to avoid and real-world applications of logarithmic equations.

Final Tips

• Make sure to use the correct property of logarithms when solving a logarithmic equation.
• Always isolate the variable $x$ when solving a logarithmic equation.
• Check for extraneous solutions when solving a logarithmic equation.
• Use the correct base for the logarithm when solving a logarithmic equation.

By following these tips and practicing solving logarithmic equations, you will become more confident and proficient in solving these types of equations.