Order The Steps To Solve The Equation $\log \left(x^2-15\right)=\log (2 X$\] From 1 To 5.1. $x^2-15=2x$2. $x^2-2x-15=0$3. $(x-5)(x+3)=0$4. $x-5=0$ Or $x+3=0$5. Potential Solutions Are $x =

Introduction

In mathematics, solving equations is a crucial skill that helps us find the value of unknown variables. In this article, we will guide you through the steps to solve the equation $\log \left(x^2-15\right)=\log (2 x)$. We will break down the solution into manageable steps, making it easier to understand and follow.

Step 1: Equating the Expressions Inside the Logarithms

The first step in solving the equation is to equate the expressions inside the logarithms. Since the logarithms are equal, we can drop the logarithms and equate the expressions inside them.

$\log \left(x^2-15\right)=\log (2 x)$

$\Rightarrow x^2-15=2x$

Step 2: Rearranging the Equation

The next step is to rearrange the equation to form a quadratic equation. We can do this by moving all the terms to one side of the equation.

$x^2-15=2x$

$\Rightarrow x^2-2x-15=0$

Step 3: Factoring the Quadratic Equation

Now that we have a quadratic equation, we can try to factor it. Factoring a quadratic equation involves finding two numbers whose product is the constant term and whose sum is the coefficient of the linear term.

$x^2-2x-15=0$

$\Rightarrow (x-5)(x+3)=0$

Step 4: Solving for x

The next step is to solve for x by setting each factor equal to zero.

$(x-5)(x+3)=0$

$\Rightarrow x-5=0$ or $x+3=0$

Step 5: Finding the Potential Solutions

The final step is to find the potential solutions by solving the two equations.

$x-5=0$

$\Rightarrow x=5$

$x+3=0$

$\Rightarrow x=-3$

Conclusion

In this article, we have walked you through the steps to solve the equation $\log \left(x^2-15\right)=\log (2 x)$. We have broken down the solution into manageable steps, making it easier to understand and follow. By following these steps, you can solve the equation and find the potential solutions.

Discussion

What do you think about the steps to solve the equation? Do you have any questions or comments? Share your thoughts in the discussion section below.

Potential Solutions

The potential solutions to the equation are $x=5$ and $x=-3$. However, we need to check if these solutions are valid by plugging them back into the original equation.

Checking the Solutions

Let's plug $x=5$ back into the original equation.

$\log \left(5^2-15\right)=\log (2 \cdot 5)$

$\log (20)=\log (10)$

Since the logarithms are equal, $x=5$ is a valid solution.

Now, let's plug $x=-3$ back into the original equation.

$\log \left((-3)^2-15\right)=\log (2 \cdot (-3))$

$\log (6)=\log (-6)$

Since the logarithms are not equal, $x=-3$ is not a valid solution.

Final Answer

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Introduction

In our previous article, we walked you through the steps to solve the equation $\log \left(x^2-15\right)=\log (2 x)$. We received many questions and comments from readers, and we're excited to share the answers with you. In this article, we'll address some of the most frequently asked questions and provide additional insights to help you better understand the solution.

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

A: The first step in solving the equation is to equate the expressions inside the logarithms. Since the logarithms are equal, we can drop the logarithms and equate the expressions inside them.

Q: Why do we need to equate the expressions inside the logarithms?

A: We need to equate the expressions inside the logarithms because the logarithms are equal. By dropping the logarithms, we can simplify the equation and make it easier to solve.

Q: What is the next step after equating the expressions inside the logarithms?

A: The next step is to rearrange the equation to form a quadratic equation. We can do this by moving all the terms to one side of the equation.

Q: How do we factor the quadratic equation?

A: We can factor the quadratic equation by finding two numbers whose product is the constant term and whose sum is the coefficient of the linear term.

Q: What are the potential solutions to the equation?

A: The potential solutions to the equation are $x=5$ and $x=-3$. However, we need to check if these solutions are valid by plugging them back into the original equation.

Q: How do we check if the solutions are valid?

A: We can check if the solutions are valid by plugging them back into the original equation. If the logarithms are equal, then the solution is valid.

Q: What is the final answer to the equation?

A: The final answer to the equation is $x=5$.

Q: Can you explain why $x=-3$ is not a valid solution?

A: $x=-3$ is not a valid solution because when we plug it back into the original equation, the logarithms are not equal. This means that $x=-3$ does not satisfy the original equation.

Q: What are some common mistakes to avoid when solving the equation?

A: Some common mistakes to avoid when solving the equation include:

• Not equating the expressions inside the logarithms
• Not rearranging the equation to form a quadratic equation
• Not factoring the quadratic equation correctly
• Not checking if the solutions are valid

Conclusion

We hope this Q&A guide has helped you better understand the solution to the equation $\log \left(x^2-15\right)=\log (2 x)$. Remember to always follow the steps carefully and check if the solutions are valid. If you have any more questions or comments, please don't hesitate to reach out.

Additional Resources

If you're looking for more resources to help you with solving equations, we recommend checking out the following:

• Khan Academy's video on solving quadratic equations
• Mathway's online calculator for solving equations
• Wolfram Alpha's online calculator for solving equations

Final Thoughts

Solving equations can be challenging, but with practice and patience, you can become proficient in solving even the most complex equations. Remember to always follow the steps carefully and check if the solutions are valid. Good luck, and happy solving!