Select The Correct Answer.What Is The Solution Set To This Equation $\log _2(x-2)+\log _2 X=3$?A. $x=3$ And $x=1$B. $x=1$C. $x=4$ And $x=-2$D. $x=4$

Introduction

Logarithmic equations can be challenging to solve, but with the right approach, they can be tackled with ease. In this article, we will explore the solution set to the equation $\log _2(x-2)+\log _2 x=3$. We will break down the solution process into manageable steps, making it easier to understand and apply.

Understanding Logarithmic Properties

Before we dive into solving the equation, it's essential to understand the properties of logarithms. The two main properties we will use are:

• Product Property: $\log _a (xy) = \log _a x + \log _a y$
• Power Property: $\log _a x^y = y \log _a x$

These properties will help us simplify the equation and make it easier to solve.

Simplifying the Equation

The given equation is $\log _2(x-2)+\log _2 x=3$. We can use the product property to combine the two logarithmic terms:

$$\log _2(x-2)+\log _2 x = \log _2((x-2)x) = 3$$

This simplifies the equation to $\log _2((x-2)x) = 3$.

Using Exponential Form

To solve the equation, we can convert it to exponential form. Since the base of the logarithm is 2, we can rewrite the equation as:

$$(x-2)x = 2^3$$

Simplifying further, we get:

$$x^2 - 2x = 8$$

Rearranging the Equation

We can rearrange the equation to form a quadratic equation:

$$x^2 - 2x - 8 = 0$$

Solving the Quadratic Equation

We can solve the quadratic equation using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 1$, $b = -2$, and $c = -8$. Plugging these values into the formula, we get:

$$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}$$

Simplifying further, we get:

$$x = \frac{2 \pm \sqrt{4 + 32}}{2}$$

$$x = \frac{2 \pm \sqrt{36}}{2}$$

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

This gives us two possible solutions:

$$x = \frac{2 + 6}{2} = 4$$

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

Checking the Solutions

We need to check if both solutions satisfy the original equation. Plugging $x = 4$ into the original equation, we get:

$$\log _2(4-2)+\log _2 4 = \log _2 2 + \log _2 4 = 1 + 2 = 3$$

This confirms that $x = 4$ is a valid solution.

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

$$\log _2(-2-2)+\log _2 (-2) = \log _2 (-4) + \log _2 (-2)$$

Since the logarithm of a negative number is undefined, $x = -2$ is not a valid solution.

Conclusion

In conclusion, the solution set to the equation $\log _2(x-2)+\log _2 x=3$ is $x = 4$. This solution satisfies the original equation and is a valid solution.

Answer

The correct answer is:

• D. $x=4$

Introduction

In our previous article, we explored the solution set to the equation $\log _2(x-2)+\log _2 x=3$. We broke down the solution process into manageable steps, making it easier to understand and apply. In this article, we will answer some frequently asked questions about solving logarithmic equations.

Q&A

Q: What is the first step in solving a logarithmic equation?

A: The first step in solving a logarithmic equation is to simplify the equation using logarithmic properties. This can involve combining logarithmic terms or rewriting the equation in exponential form.

Q: How do I convert a logarithmic equation to exponential form?

A: To convert a logarithmic equation to exponential form, you can use the definition of a logarithm: $\log _a x = y$ is equivalent to $a^y = x$. For example, the equation $\log _2 x = 3$ can be rewritten as $2^3 = x$, which simplifies to $x = 8$.

Q: What is the quadratic formula, and how do I use it to solve a quadratic equation?

A: The quadratic formula is a mathematical formula that can be used to solve quadratic equations. It is given by:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

To use the quadratic formula, you need to identify the values of $a$, $b$, and $c$ in the quadratic equation. Then, you can plug these values into the formula to find the solutions.

Q: How do I check if a solution satisfies the original equation?

A: To check if a solution satisfies the original equation, you need to plug the solution back into the original equation and simplify. If the simplified expression is true, then the solution is valid. If the simplified expression is false, then the solution is not valid.

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

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

• Not simplifying the equation using logarithmic properties
• Not converting the equation to exponential form
• Not using the quadratic formula to solve quadratic equations
• Not checking if the solution satisfies the original equation

Tips and Tricks

Tip 1: Simplify the equation using logarithmic properties

Before solving a logarithmic equation, try to simplify it using logarithmic properties. This can involve combining logarithmic terms or rewriting the equation in exponential form.

Tip 2: Use the quadratic formula to solve quadratic equations

If the equation is quadratic, try using the quadratic formula to solve it. This can save you time and effort in the long run.

Tip 3: Check if the solution satisfies the original equation

Before accepting a solution, make sure to check if it satisfies the original equation. This can help you avoid mistakes and ensure that your solution is correct.

Conclusion

Solving logarithmic equations can be challenging, but with the right approach, it can be done with ease. By simplifying the equation using logarithmic properties, converting it to exponential form, and using the quadratic formula to solve quadratic equations, you can find the solution set to the equation. Remember to check if the solution satisfies the original equation and avoid common mistakes. With practice and patience, you can become proficient in solving logarithmic equations.

Frequently Asked Questions

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

A: A logarithmic equation is an equation that involves a logarithm, while an exponential equation is an equation that involves an exponential expression. For example, the equation $\log _2 x = 3$ is a logarithmic equation, while the equation $2^x = 8$ is an exponential equation.

Q: How do I solve an exponential equation?

A: To solve an exponential equation, you can use the definition of an exponential function: $a^x = y$ is equivalent to $x = \log _a y$. For example, the equation $2^x = 8$ can be rewritten as $x = \log _2 8$, which simplifies to $x = 3$.

Q: What is the relationship between logarithmic and exponential functions?

A: Logarithmic and exponential functions are inverse functions. This means that if $f(x) = \log _a x$, then $f^{-1}(x) = a^x$. For example, if $f(x) = \log _2 x$, then $f^{-1}(x) = 2^x$.

Additional Resources

For more information on solving logarithmic equations, check out the following resources:

• Khan Academy: Logarithmic Equations
• Mathway: Logarithmic Equations
• Wolfram Alpha: Logarithmic Equations

By following these tips and resources, you can become proficient in solving logarithmic equations and tackle even the most challenging problems with confidence.