What Is The Solution To This Equation? Log ⁡ 8 16 + 2 Log ⁡ 8 X = 2 \log_8 16 + 2\log_8 X = 2 Lo G 8 ​ 16 + 2 Lo G 8 ​ X = 2 A. X = 0 X = 0 X = 0 B. X = 2 X = 2 X = 2 C. X = − 2 X = -2 X = − 2 D. X = 4 X = 4 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 focus on solving a specific logarithmic equation, $\log_8 16 + 2\log_8 x = 2$, and explore the different solution options. We will break down the solution process into manageable steps, making it easier for readers to understand and apply the concepts.

Understanding Logarithmic Equations

Before we dive into the solution, let's take a moment to understand the basics of logarithmic equations. A logarithmic equation is an equation that involves a logarithm, which is the inverse operation of exponentiation. In other words, if $a^b = c$, then $\log_a c = b$. Logarithmic equations can be solved using various techniques, including the use of logarithmic properties and the change of base formula.

The Given Equation

The given equation is $\log_8 16 + 2\log_8 x = 2$. To solve this equation, we need to isolate the variable $x$. Let's start by simplifying the equation using the properties of logarithms.

Simplifying the Equation

We can start by simplifying the first term, $\log_8 16$. Since $16 = 2^4$ and $8 = 2^3$, we can rewrite the first term as:

$$\log_8 16 = \log_{2^3} 2^4 = \frac{4}{3}$$

Now, let's substitute this value back into the original equation:

$$\frac{4}{3} + 2\log_8 x = 2$$

Isolating the Variable

Next, let's isolate the variable $x$ by subtracting $\frac{4}{3}$ from both sides of the equation:

$$2\log_8 x = 2 - \frac{4}{3}$$

Simplifying the right-hand side, we get:

$$2\log_8 x = \frac{2}{3}$$

Using Logarithmic Properties

Now, let's use the property of logarithms that states $\log_a b^c = c\log_a b$. We can rewrite the equation as:

$$\log_8 x^2 = \frac{1}{3}$$

Solving for x

Finally, let's solve for $x$ by exponentiating both sides of the equation:

$$x^2 = 8^{\frac{1}{3}}$$

Simplifying the right-hand side, we get:

$$x^2 = 2$$

Taking the square root of both sides, we get:

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

Checking the Solutions

Now that we have found the solutions, let's check them to see if they satisfy the original equation. Plugging in $x = \sqrt{2}$, we get:

$$\log_8 16 + 2\log_8 \sqrt{2} = 2$$

Simplifying the left-hand side, we get:

$$\log_8 16 + \log_8 2 = 2$$

Using the property of logarithms that states $\log_a b + \log_a c = \log_a bc$, we can rewrite the equation as:

$$\log_8 32 = 2$$

Exponentiating both sides, we get:

$$8^2 = 32$$

Which is true. Therefore, $x = \sqrt{2}$ is a valid solution.

Plugging in $x = -\sqrt{2}$, we get:

$$\log_8 16 + 2\log_8 -\sqrt{2} = 2$$

Simplifying the left-hand side, we get:

$$\log_8 16 - \log_8 2 = 2$$

Using the property of logarithms that states $\log_a b - \log_a c = \log_a \frac{b}{c}$, we can rewrite the equation as:

$$\log_8 \frac{16}{2} = 2$$

Simplifying the left-hand side, we get:

$$\log_8 8 = 2$$

Exponentiating both sides, we get:

$$8^2 = 8$$

Which is not true. Therefore, $x = -\sqrt{2}$ is not a valid solution.

Conclusion

In conclusion, the solution to the equation $\log_8 16 + 2\log_8 x = 2$ is $x = \sqrt{2}$. This solution satisfies the original equation and is therefore valid.

Answer

The correct answer is:

• $\boxed{\sqrt{2}}$

$$$$$$

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 the properties of logarithms. This may involve combining logarithms, using the change of base formula, or applying other logarithmic properties to isolate the variable.

Q: How do I simplify a logarithmic equation?

A: To simplify a logarithmic equation, you can use the following properties of logarithms:

• $\log_a b^c = c\log_a b$
• $\log_a b + \log_a c = \log_a bc$
• $\log_a b - \log_a c = \log_a \frac{b}{c}$

You can also use the change of base formula to rewrite a logarithmic equation in terms of a different base.

Q: What is the change of base formula?

A: The change of base formula is a mathematical formula that allows you to rewrite a logarithmic equation in terms of a different base. The formula is:

$$\log_a b = \frac{\log_c b}{\log_c a}$$

where $a$, $b$, and $c$ are positive real numbers and $c \neq 1$.

Q: How do I isolate the variable in a logarithmic equation?

A: To isolate the variable in a logarithmic equation, you can use the following steps:

1. Simplify the equation using the properties of logarithms.
2. Use algebraic manipulations to isolate the variable.
3. Exponentiate both sides of the equation to solve for the variable.

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

A: A logarithmic equation is an equation that involves a logarithm, which is the inverse operation of exponentiation. An exponential equation, on the other hand, is an equation that involves an exponent, which is the inverse operation of a logarithm.

For example, the equation $\log_2 x = 3$ is a logarithmic equation, while the equation $2^x = 8$ is an exponential equation.

Q: Can I use a calculator to solve a logarithmic equation?

A: Yes, you can use a calculator to solve a logarithmic equation. However, it's always a good idea to check your work by plugging the solution back into the original equation to make sure it's true.

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

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

• Forgetting to simplify the equation using the properties of logarithms.
• Not isolating the variable correctly.
• Not checking the solution by plugging it back into the original equation.
• Using the wrong base or exponent.

By avoiding these common mistakes, you can ensure that your solutions are accurate and reliable.

Q: Can I use logarithmic equations to model real-world problems?

A: Yes, logarithmic equations can be used to model a wide range of real-world problems, including population growth, chemical reactions, and financial transactions.

For example, the equation $\log_2 P = t$ can be used to model the growth of a population over time, where $P$ is the population size and $t$ is the time.

Q: What are some applications of logarithmic equations in science and engineering?

A: Logarithmic equations have a wide range of applications in science and engineering, including:

• Population growth and modeling
• Chemical reactions and kinetics
• Financial transactions and economics
• Signal processing and communication systems
• Image and video processing

By understanding and applying logarithmic equations, you can gain a deeper insight into the underlying principles and mechanisms that govern these complex systems.

Conclusion

In conclusion, logarithmic equations are a powerful tool for modeling and solving a wide range of real-world problems. By understanding the properties and applications of logarithmic equations, you can gain a deeper insight into the underlying principles and mechanisms that govern these complex systems. Whether you're a student, a researcher, or a practitioner, logarithmic equations are an essential part of your toolkit.