Order The Steps To Solve The Equation $\log _3(x+2)=\log _3\left(2 X^2-1\right$\] From 1 To 6.1. $3^{\log _3(x+2)}=3^{\log _3\left(2 X^2-1\right)}$2. $x+2=2 X^2-1$3. $0=2 X^2-x-3$4. $0=(2 X-3)(x+1$\]5. $2

Introduction

Logarithmic equations can be challenging to solve, but with a clear understanding of the steps involved, they can be tackled with ease. In this article, we will guide you through the process of solving the equation $\log _3(x+2)=\log _3\left(2 x^2-1\right)$, breaking it down into manageable steps. We will order the steps from 1 to 6.1, providing a clear and concise explanation of each step.

Step 1: Apply the One-to-One Property of Logarithms

The first step in solving the equation is to apply the one-to-one property of logarithms, which states that if $\log _a(x)=\log _a(y)$, then $x=y$. In this case, we have:

$$\log _3(x+2)=\log _3\left(2 x^2-1\right)$$

Using the one-to-one property, we can rewrite the equation as:

$$x+2=2 x^2-1$$

Step 2: Simplify the Equation

The next step is to simplify the equation by combining like terms. We can do this by moving all the terms to one side of the equation:

$$x+2=2 x^2-1$$

Subtracting $x$ from both sides gives us:

$$2=x^2-1$$

Adding 1 to both sides gives us:

$$3=x^2$$

Step 3: Factor the Quadratic Equation

The equation $x^2=3$ is a quadratic equation, and we can factor it as:

$$x^2-3=0$$

However, this is not the correct factorization. We need to factor the original equation $x+2=2 x^2-1$ instead. We can do this by factoring the quadratic expression $2 x^2-x-3$:

$$2 x^2-x-3=(2 x-3)(x+1)$$

Step 4: Set Each Factor Equal to Zero

Now that we have factored the quadratic expression, we can set each factor equal to zero:

$$(2 x-3)(x+1)=0$$

This gives us two possible solutions:

$$2 x-3=0 \quad \text{or} \quad x+1=0$$

Step 5: Solve for x

We can solve for $x$ by isolating it in each equation:

$$2 x-3=0 \quad \Rightarrow \quad 2 x=3 \quad \Rightarrow \quad x=\frac{3}{2}$$

$$x+1=0 \quad \Rightarrow \quad x=-1$$

Step 6.1: Check the Solutions

Finally, we need to check our solutions to make sure they are valid. We can do this by plugging each solution back into the original equation:

$$\log _3\left(\frac{3}{2}+2\right)=\log _3\left(2 \left(\frac{3}{2}\right)^2-1\right)$$

$$\log _3\left(\frac{7}{2}\right)=\log _3\left(\frac{7}{2}\right)$$

This shows that $x=\frac{3}{2}$ is a valid solution.

$$\log _3\left(-1+2\right)=\log _3\left(2 (-1)^2-1\right)$$

$$\log _3(1)=\log _3(1)$$

This shows that $x=-1$ is also a valid solution.

Conclusion

$$$$$$$$

Q: What is the one-to-one property of logarithms?

A: The one-to-one property of logarithms states that if $\log _a(x)=\log _a(y)$, then $x=y$. This means that if the logarithms of two expressions are equal, then the expressions themselves must be equal.

Q: How do I apply the one-to-one property of logarithms to solve a logarithmic equation?

A: To apply the one-to-one property of logarithms, you need to rewrite the equation in exponential form. For example, if you have the equation $\log _3(x+2)=\log _3\left(2 x^2-1\right)$, you can rewrite it as $3^{\log _3(x+2)}=3^{\log _3\left(2 x^2-1\right)}$. Then, you can simplify the equation by canceling out the logarithms.

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

A: A logarithmic equation is an equation that involves a logarithm, such as $\log _3(x+2)=\log _3\left(2 x^2-1\right)$. A quadratic equation, on the other hand, is an equation that involves a quadratic expression, such as $x^2-3=0$. While both types of equations can be solved using algebraic techniques, they require different approaches.

Q: How do I factor a quadratic expression?

A: To factor a quadratic expression, you need to find two numbers whose product is equal to the constant term and whose sum is equal to the coefficient of the linear term. For example, if you have the quadratic expression $x^2-3$, you can factor it as $(x- \sqrt{3})(x+ \sqrt{3})$. However, if you have a quadratic expression like $2 x^2-x-3$, you need to factor it as $(2 x-3)(x+1)$.

Q: What is the difference between a solution and a valid solution?

A: A solution is a value that satisfies the equation, but it may not be a valid solution. A valid solution is a value that satisfies the equation and also makes sense in the context of the problem. For example, if you have the equation $\log _3(x+2)=\log _3\left(2 x^2-1\right)$, the solution $x=-1$ satisfies the equation, but it is not a valid solution because it makes the expression $x+2$ negative.

Q: How do I check my solutions to make sure they are valid?

A: To check your solutions, you need to plug each solution back into the original equation and make sure it is true. For example, if you have the solution $x=\frac{3}{2}$, you can plug it back into the equation $\log _3(x+2)=\log _3\left(2 x^2-1\right)$ and make sure it is true.

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 one-to-one property of logarithms
• Not simplifying the equation
• Not factoring the quadratic expression correctly
• Not checking the solutions to make sure they are valid
• Not considering the domain of the logarithmic function

Conclusion

In this article, we have answered some frequently asked questions about solving logarithmic equations. We have covered topics such as the one-to-one property of logarithms, applying the one-to-one property, factoring quadratic expressions, and checking solutions. By following these tips and avoiding common mistakes, you can become proficient in solving logarithmic equations.