What Are The Solutions To Log ⁡ 6 ( X 2 + 8 ) = 1 + Log ⁡ 6 ( X \log_6\left(x^2+8\right)=1+\log_6(x Lo G 6 ​ ( X 2 + 8 ) = 1 + Lo G 6 ​ ( X ]?A. X = − 2 X=-2 X = − 2 And X = − 4 X=-4 X = − 4 B. X = − 1 X=-1 X = − 1 And X = 8 X=8 X = 8 C. X = 1 X=1 X = 1 And X = − 8 X=-8 X = − 8 D. X = 2 X=2 X = 2 And X = 4 X=4 X = 4

What are the solutions to $\log_6\left(x^2+8\right)=1+\log_6(x)$?

Understanding the Problem

The given equation involves logarithms and a quadratic expression. To solve for the value of $x$, we need to manipulate the equation using logarithmic properties and algebraic techniques. The equation is $\log_6\left(x^2+8\right)=1+\log_6(x)$. Our goal is to find the values of $x$ that satisfy this equation.

Using Logarithmic Properties

We can start by using the property of logarithms that states $\log_a(b) + \log_a(c) = \log_a(bc)$. Applying this property to the given equation, we get:

$$\log_6\left(x^2+8\right) = \log_6(x) + \log_6(6)$$

Since $\log_a(a) = 1$, we can rewrite the equation as:

$$\log_6\left(x^2+8\right) = \log_6(x) + 1$$

Simplifying the Equation

Now, we can use the property of logarithms that states $\log_a(b) = c$ is equivalent to $b = a^c$. Applying this property to the given equation, we get:

$$x^2 + 8 = 6^{\log_6(x) + 1}$$

We can simplify the right-hand side of the equation using the property of exponents that states $a^{b+c} = a^b \cdot a^c$. Applying this property, we get:

$$x^2 + 8 = 6^{\log_6(x)} \cdot 6^1$$

Since $6^{\log_6(x)} = x$, we can rewrite the equation as:

$$x^2 + 8 = x \cdot 6$$

Solving the Quadratic Equation

Now, we have a quadratic equation in the form $x^2 + 8 = 6x$. We can rearrange the equation to get:

$$x^2 - 6x + 8 = 0$$

This is a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -6$, and $c = 8$. We can solve this equation using the quadratic formula:

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

Plugging in the values of $a$, $b$, and $c$, we get:

$$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)}$$

Simplifying the expression, we get:

$$x = \frac{6 \pm \sqrt{36 - 32}}{2}$$

$$x = \frac{6 \pm \sqrt{4}}{2}$$

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

This gives us two possible values for $x$:

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

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

Checking the Solutions

We need to check if these values of $x$ satisfy the original equation. Plugging in $x = 4$, we get:

$$\log_6\left(4^2 + 8\right) = \log_6(4) + 1$$

$$\log_6\left(16 + 8\right) = \log_6(4) + 1$$

$$\log_6\left(24\right) = \log_6(4) + 1$$

Since $\log_6(24) \neq \log_6(4) + 1$, $x = 4$ is not a solution.

Plugging in $x = 2$, we get:

$$\log_6\left(2^2 + 8\right) = \log_6(2) + 1$$

$$\log_6\left(4 + 8\right) = \log_6(2) + 1$$

$$\log_6\left(12\right) = \log_6(2) + 1$$

Since $\log_6(12) = \log_6(2) + 1$, $x = 2$ is a solution.

Conclusion

The solution to the equation $\log_6\left(x^2+8\right)=1+\log_6(x)$ is $x = 2$. This value satisfies the original equation and is the only solution.

Answer

The correct answer is D. $x=2$ and $x=4$
Q&A: Understanding the Solutions to $\log_6\left(x^2+8\right)=1+\log_6(x)$

Q: What is the main concept behind solving the equation $\log_6\left(x^2+8\right)=1+\log_6(x)$?

A: The main concept behind solving this equation is using logarithmic properties and algebraic techniques to manipulate the equation and isolate the variable $x$.

Q: How do you use logarithmic properties to simplify the equation?

A: We can use the property of logarithms that states $\log_a(b) + \log_a(c) = \log_a(bc)$. Applying this property to the given equation, we get:

$$\log_6\left(x^2+8\right) = \log_6(x) + \log_6(6)$$

Since $\log_a(a) = 1$, we can rewrite the equation as:

$$\log_6\left(x^2+8\right) = \log_6(x) + 1$$

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

A: We can use the property of logarithms that states $\log_a(b) = c$ is equivalent to $b = a^c$. Applying this property to the given equation, we get:

$$x^2 + 8 = 6^{\log_6(x) + 1}$$

We can simplify the right-hand side of the equation using the property of exponents that states $a^{b+c} = a^b \cdot a^c$. Applying this property, we get:

$$x^2 + 8 = 6^{\log_6(x)} \cdot 6^1$$

Since $6^{\log_6(x)} = x$, we can rewrite the equation as:

$$x^2 + 8 = x \cdot 6$$

Q: How do you solve the quadratic equation $x^2 - 6x + 8 = 0$?

A: We can use the quadratic formula:

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

Plugging in the values of $a$, $b$, and $c$, we get:

$$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)}$$

Simplifying the expression, we get:

$$x = \frac{6 \pm \sqrt{36 - 32}}{2}$$

$$x = \frac{6 \pm \sqrt{4}}{2}$$

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

This gives us two possible values for $x$:

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

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

Q: How do you check if these values of $x$ satisfy the original equation?

A: We need to plug in the values of $x$ into the original equation and check if it is true. For $x = 4$, we get:

$$\log_6\left(4^2 + 8\right) = \log_6(4) + 1$$

$$\log_6\left(16 + 8\right) = \log_6(4) + 1$$

$$\log_6\left(24\right) = \log_6(4) + 1$$

Since $\log_6(24) \neq \log_6(4) + 1$, $x = 4$ is not a solution.

For $x = 2$, we get:

$$\log_6\left(2^2 + 8\right) = \log_6(2) + 1$$

$$\log_6\left(4 + 8\right) = \log_6(2) + 1$$

$$\log_6\left(12\right) = \log_6(2) + 1$$

Since $\log_6(12) = \log_6(2) + 1$, $x = 2$ is a solution.

Q: What is the final answer to the equation $\log_6\left(x^2+8\right)=1+\log_6(x)$?

A: The final answer is $x = 2$. This value satisfies the original equation and is the only solution.

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

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

• Not using the correct logarithmic properties
• Not simplifying the equation correctly
• Not checking if the solutions satisfy the original equation
• Not using the correct quadratic formula

By avoiding these mistakes, you can ensure that you are solving the equation correctly and finding the correct solution.