Select The Correct Answer.What Is The Approximate Solution To This Equation? 19 + 2 Ln ⁡ X = 25 19 + 2 \ln X = 25 19 + 2 Ln X = 25 A. X ≈ 1.93 X \approx 1.93 X ≈ 1.93 B. X ≈ 0.05 X \approx 0.05 X ≈ 0.05 C. X ≈ 3 X \approx 3 X ≈ 3 D. X ≈ 20.09 X \approx 20.09 X ≈ 20.09

Introduction

In this article, we will be solving the equation $19 + 2 \ln x = 25$ to find the approximate solution for $x$. This equation involves a natural logarithm, and we will use algebraic manipulations to isolate the variable $x$. We will also provide a step-by-step guide on how to solve this equation, making it easy to understand for readers who are new to solving logarithmic equations.

Understanding the Equation

The given equation is $19 + 2 \ln x = 25$. This equation involves a natural logarithm, which is denoted by $\ln x$. The natural logarithm is the inverse of the exponential function, and it is used to solve equations that involve exponential functions.

Step 1: Isolate the Natural Logarithm

To solve this equation, we need to isolate the natural logarithm. We can do this by subtracting 19 from both sides of the equation:

$$2 \ln x = 25 - 19$$

This simplifies to:

$$2 \ln x = 6$$

Step 2: Divide Both Sides by 2

Next, we need to divide both sides of the equation by 2 to isolate the natural logarithm:

$$\ln x = \frac{6}{2}$$

This simplifies to:

$$\ln x = 3$$

Step 3: Exponentiate Both Sides

To solve for $x$, we need to exponentiate both sides of the equation. Since the natural logarithm is the inverse of the exponential function, we can use the exponential function to solve for $x$:

$$x = e^3$$

Step 4: Approximate the Value of $x$

To find the approximate value of $x$, we can use a calculator to evaluate the exponential function:

$$x \approx e^3 \approx 20.08554$$

Conclusion

In this article, we solved the equation $19 + 2 \ln x = 25$ to find the approximate solution for $x$. We used algebraic manipulations to isolate the natural logarithm and then exponentiated both sides of the equation to solve for $x$. The approximate value of $x$ is $20.08554$, which is closest to option D.

Answer

The correct answer is:

• D. $x \approx 20.09$

Discussion

This equation involves a natural logarithm, and we used algebraic manipulations to isolate the variable $x$. We also used the exponential function to solve for $x$. This equation is a good example of how to solve logarithmic equations, and it can be used as a reference for readers who are new to solving logarithmic equations.

Tips and Tricks

• When solving logarithmic equations, it is essential to isolate the natural logarithm first.
• Use algebraic manipulations to simplify the equation and make it easier to solve.
• Use the exponential function to solve for the variable.
• Approximate the value of the variable using a calculator.

Related Topics

• Solving exponential equations
• Solving logarithmic equations
• Algebraic manipulations
• Exponential functions

Conclusion

$$$$$$$$$$ $$

Introduction

In our previous article, we solved the equation $19 + 2 \ln x = 25$ to find the approximate solution for $x$. We used algebraic manipulations to isolate the natural logarithm and then exponentiated both sides of the equation to solve for $x$. In this article, we will answer some frequently asked questions about solving this equation.

Q: What is the first step in solving the equation $19 + 2 \ln x = 25$?

A: The first step in solving the equation $19 + 2 \ln x = 25$ is to isolate the natural logarithm. We can do this by subtracting 19 from both sides of the equation.

Q: How do I isolate the natural logarithm in the equation $19 + 2 \ln x = 25$?

A: To isolate the natural logarithm, we need to subtract 19 from both sides of the equation. This will give us:

$$2 \ln x = 25 - 19$$

This simplifies to:

$$2 \ln x = 6$$

Q: What is the next step in solving the equation $19 + 2 \ln x = 25$?

A: The next step in solving the equation $19 + 2 \ln x = 25$ is to divide both sides of the equation by 2 to isolate the natural logarithm.

Q: How do I divide both sides of the equation $2 \ln x = 6$ by 2?

A: To divide both sides of the equation $2 \ln x = 6$ by 2, we can simply divide both sides of the equation by 2:

$$\ln x = \frac{6}{2}$$

This simplifies to:

$$\ln x = 3$$

Q: What is the final step in solving the equation $19 + 2 \ln x = 25$?

A: The final step in solving the equation $19 + 2 \ln x = 25$ is to exponentiate both sides of the equation to solve for $x$.

Q: How do I exponentiate both sides of the equation $\ln x = 3$?

A: To exponentiate both sides of the equation $\ln x = 3$, we can use the exponential function:

$$x = e^3$$

Q: What is the approximate value of $x$ in the equation $x = e^3$?

A: To find the approximate value of $x$ in the equation $x = e^3$, we can use a calculator to evaluate the exponential function:

$$x \approx e^3 \approx 20.08554$$

Q: Which option is closest to the approximate value of $x$?

A: The approximate value of $x$ is $20.08554$, which is closest to option D.

Conclusion

In this article, we answered some frequently asked questions about solving the equation $19 + 2 \ln x = 25$. We provided step-by-step instructions on how to isolate the natural logarithm, divide both sides of the equation by 2, and exponentiate both sides of the equation to solve for $x$. We also provided the approximate value of $x$ and the closest option.

Tips and Tricks

• When solving logarithmic equations, it is essential to isolate the natural logarithm first.
• Use algebraic manipulations to simplify the equation and make it easier to solve.
• Use the exponential function to solve for the variable.
• Approximate the value of the variable using a calculator.

Related Topics

• Solving exponential equations
• Solving logarithmic equations
• Algebraic manipulations
• Exponential functions

Conclusion

In this article, we provided a Q&A section on solving the equation $19 + 2 \ln x = 25$. We answered some frequently asked questions and provided step-by-step instructions on how to solve the equation. We also provided the approximate value of $x$ and the closest option. This article is a good reference for readers who are new to solving logarithmic equations.