What Are The Potential Solutions Of Log ⁡ 4 X + Log ⁡ 4 ( X + 6 ) = 2 \log _4 X+\log _4(x+6)=2 Lo G 4 ​ X + Lo G 4 ​ ( X + 6 ) = 2 ?A. X = − 2 X=-2 X = − 2 And X = − 8 X=-8 X = − 8 B. X = − 2 X=-2 X = − 2 And X = 8 X=8 X = 8 C. X = 2 X=2 X = 2 And X = − 8 X=-8 X = − 8 D. X = 2 X=2 X = 2 And X = 8 X=8 X = 8

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Introduction

In this article, we will explore the potential solutions of the logarithmic equation $\log _4 x+\log _4(x+6)=2$. This equation involves logarithms with base 4, and our goal is to find the values of $x$ that satisfy this equation. We will use various mathematical techniques, including properties of logarithms and algebraic manipulations, to solve this equation and determine the potential solutions.

Understanding the Equation

The given equation is $\log _4 x+\log _4(x+6)=2$. This equation involves two logarithmic terms, each with base 4. The first term is $\log _4 x$, and the second term is $\log _4(x+6)$. The equation states that the sum of these two logarithmic terms is equal to 2.

Using Properties of Logarithms

To solve this equation, we can use the property of logarithms that states $\log _a b + \log _a c = \log _a (bc)$. This property allows us to combine the two logarithmic terms into a single logarithmic term.

Using this property, we can rewrite the equation as:

$$\log _4 (x(x+6)) = 2$$

Simplifying the Equation

Now that we have combined the two logarithmic terms, we can simplify the equation further. We can rewrite the equation as:

$$x(x+6) = 4^2$$

Expanding and Simplifying

We can expand the left-hand side of the equation by multiplying the two terms:

$$x^2 + 6x = 16$$

Rearranging the Equation

We can rearrange the equation to form a quadratic equation:

$$x^2 + 6x - 16 = 0$$

Solving the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

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

In this case, $a = 1$, $b = 6$, and $c = -16$. Plugging these values into the quadratic formula, we get:

$$x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-16)}}{2(1)}$$

Simplifying the Quadratic Formula

We can simplify the quadratic formula by evaluating the expression inside the square root:

$$x = \frac{-6 \pm \sqrt{36 + 64}}{2}$$

$$x = \frac{-6 \pm \sqrt{100}}{2}$$

Evaluating the Square Root

We can evaluate the square root of 100:

$$x = \frac{-6 \pm 10}{2}$$

Finding the Potential Solutions

We can find the potential solutions by evaluating the two possible values of $x$:

$$x = \frac{-6 + 10}{2} = 2$$

$$x = \frac{-6 - 10}{2} = -8$$

Conclusion

In this article, we have explored the potential solutions of the logarithmic equation $\log _4 x+\log _4(x+6)=2$. We have used various mathematical techniques, including properties of logarithms and algebraic manipulations, to solve this equation and determine the potential solutions. The potential solutions are $x = 2$ and $x = -8$.

Final Answer

The final answer is $\boxed{C. x=2 and x=-8}$

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Introduction

In our previous article, we explored the potential solutions of the logarithmic equation $\log _4 x+\log _4(x+6)=2$. We used various mathematical techniques, including properties of logarithms and algebraic manipulations, to solve this equation and determine the potential solutions. In this article, we will answer some frequently asked questions (FAQs) about this equation.

Q: What is the base of the logarithm in the equation $\log _4 x+\log _4(x+6)=2$?

A: The base of the logarithm in the equation $\log _4 x+\log _4(x+6)=2$ is 4.

Q: What is the property of logarithms used to solve the equation $\log _4 x+\log _4(x+6)=2$?

A: The property of logarithms used to solve the equation $\log _4 x+\log _4(x+6)=2$ is $\log _a b + \log _a c = \log _a (bc)$.

Q: How do we simplify the equation $\log _4 (x(x+6)) = 2$?

A: We simplify the equation $\log _4 (x(x+6)) = 2$ by rewriting it as $x(x+6) = 4^2$.

Q: What is the quadratic equation formed by rearranging the equation $x(x+6) = 16$?

A: The quadratic equation formed by rearranging the equation $x(x+6) = 16$ is $x^2 + 6x - 16 = 0$.

Q: How do we solve the quadratic equation $x^2 + 6x - 16 = 0$?

A: We solve the quadratic equation $x^2 + 6x - 16 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Q: What are the potential solutions of the equation $\log _4 x+\log _4(x+6)=2$?

A: The potential solutions of the equation $\log _4 x+\log _4(x+6)=2$ are $x = 2$ and $x = -8$.

Q: Why are the potential solutions $x = 2$ and $x = -8$?

A: The potential solutions $x = 2$ and $x = -8$ are the values of $x$ that satisfy the equation $\log _4 x+\log _4(x+6)=2$. These values are obtained by solving the quadratic equation $x^2 + 6x - 16 = 0$ using the quadratic formula.

Q: What is the final answer to the equation $\log _4 x+\log _4(x+6)=2$?

A: The final answer to the equation $\log _4 x+\log _4(x+6)=2$ is $\boxed{C. x=2 and x=-8}$.

Conclusion

In this article, we have answered some frequently asked questions (FAQs) about the logarithmic equation $\log _4 x+\log _4(x+6)=2$. We have provided detailed explanations and examples to help clarify the concepts and techniques used to solve this equation. We hope that this article has been helpful in understanding the potential solutions of this equation.