Solve The Following Logarithmic Equation:$\[ 3 + 8 \log _{\frac{1}{k}} \left( \sqrt{8 + 4 \sqrt{3}} - \sqrt{8 - 4 \sqrt{3}} \right) = 0, \quad K \ \textgreater \ 0, \quad K \neq 1 \\]

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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 that involves a base of $\frac{1}{k}$, where $k$ is a positive number greater than 1.

Understanding the Equation

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The given equation is:

${ 3 + 8 \log _{\frac{1}{k}} \left( \sqrt{8 + 4 \sqrt{3}} - \sqrt{8 - 4 \sqrt{3}} \right) = 0, \quad k \ \textgreater \ 0, \quad k \neq 1 }$

This equation involves a logarithm with a base of $\frac{1}{k}$, which can be rewritten as a logarithm with a base of $k$. To simplify the equation, we can use the property of logarithms that states $\log_b a = -\log_{\frac{1}{b}} a$.

Simplifying the Equation

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Using the property of logarithms, we can rewrite the equation as:

${ 3 - 8 \log_k \left( \sqrt{8 + 4 \sqrt{3}} - \sqrt{8 - 4 \sqrt{3}} \right) = 0 }$

Now, we can simplify the expression inside the logarithm by factoring the square roots:

${ \sqrt{8 + 4 \sqrt{3}} - \sqrt{8 - 4 \sqrt{3}} = \sqrt{4(2 + \sqrt{3})} - \sqrt{4(2 - \sqrt{3})} }$

${ = 2\sqrt{2 + \sqrt{3}} - 2\sqrt{2 - \sqrt{3}} }$

Using the Difference of Squares Formula

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We can use the difference of squares formula to simplify the expression further:

${ (a + b)(a - b) = a^2 - b^2 }$

In this case, we can let $a = 2\sqrt{2 + \sqrt{3}}$ and $b = 2\sqrt{2 - \sqrt{3}}$.

Simplifying the Expression

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Using the difference of squares formula, we can simplify the expression as:

${ 2\sqrt{2 + \sqrt{3}} - 2\sqrt{2 - \sqrt{3}} = \sqrt{(2\sqrt{2 + \sqrt{3}})^2 - (2\sqrt{2 - \sqrt{3}})^2} }$

${ = \sqrt{4(2 + \sqrt{3}) - 4(2 - \sqrt{3})} }$

${ = \sqrt{8 + 4\sqrt{3} - 8 + 4\sqrt{3}} }$

${ = \sqrt{8\sqrt{3}} }$

${ = 2\sqrt{2\sqrt{3}} }$

Substituting the Simplified Expression

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Now, we can substitute the simplified expression back into the original equation:

${ 3 - 8 \log_k (2\sqrt{2\sqrt{3}}) = 0 }$

Using the Property of Logarithms

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We can use the property of logarithms that states $\log_b a^c = c \log_b a$ to simplify the equation further:

${ 3 - 8 \log_k (2\sqrt{2\sqrt{3}}) = 0 }$

${ 3 - 8 \log_k (2^2 \cdot \sqrt{2\sqrt{3}}) = 0 }$

${ 3 - 8 \log_k (2^2) - 8 \log_k (\sqrt{2\sqrt{3}}) = 0 }$

Simplifying the Equation

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Using the property of logarithms, we can simplify the equation as:

${ 3 - 8 \log_k (2^2) - 8 \log_k (\sqrt{2\sqrt{3}}) = 0 }$

${ 3 - 16 \log_k (2) - 8 \log_k (\sqrt{2\sqrt{3}}) = 0 }$

Using the Property of Logarithms

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We can use the property of logarithms that states $\log_b a^c = c \log_b a$ to simplify the equation further:

${ 3 - 16 \log_k (2) - 8 \log_k (\sqrt{2\sqrt{3}}) = 0 }$

${ 3 - 16 \log_k (2) - 4 \log_k (2\sqrt{3}) = 0 }$

Simplifying the Equation

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Using the property of logarithms, we can simplify the equation as:

${ 3 - 16 \log_k (2) - 4 \log_k (2\sqrt{3}) = 0 }$

${ 3 - 16 \log_k (2) - 4 \log_k (2) - 4 \log_k (\sqrt{3}) = 0 }$

Using the Property of Logarithms

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We can use the property of logarithms that states $\log_b a^c = c \log_b a$ to simplify the equation further:

${ 3 - 16 \log_k (2) - 4 \log_k (2) - 4 \log_k (\sqrt{3}) = 0 }$

${ 3 - 20 \log_k (2) - 4 \log_k (\sqrt{3}) = 0 }$

Simplifying the Equation

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Using the property of logarithms, we can simplify the equation as:

${ 3 - 20 \log_k (2) - 4 \log_k (\sqrt{3}) = 0 }$

${ 3 - 20 \log_k (2) - 2 \log_k (3) = 0 }$

Using the Property of Logarithms

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We can use the property of logarithms that states $\log_b a^c = c \log_b a$ to simplify the equation further:

${ 3 - 20 \log_k (2) - 2 \log_k (3) = 0 }$

${ 3 - 20 \log_k (2) - 2 \log_k (3) = 0 }$

Solving for k

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Now, we can solve for $k$ by isolating the logarithmic terms:

${ 3 - 20 \log_k (2) - 2 \log_k (3) = 0 }$

${ 20 \log_k (2) + 2 \log_k (3) = 3 }$

${ \log_k (2^{20}) + \log_k (3^2) = 3 }$

${ \log_k (2^{20} \cdot 3^2) = 3 }$

${ 2^{20} \cdot 3^2 = k^3 }$

Taking the Cube Root of Both Sides

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Taking the cube root of both sides, we get:

${ \sqrt[3]{2^{20} \cdot 3^2} = k }$

${ k = \sqrt[3]{2^{20} \cdot 3^2} }$

Simplifying the Expression

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Simplifying the expression, we get:

${ k = \sqrt[3]{2^{20} \cdot 3^2} }$

${ k = \sqrt[3]{2^{20} \cdot 9} }$

${ k = \sqrt[3]{2^{20} \cdot 3^2} }$

${ k = \sqrt[3]{(2^4)^5 \cdot 3^2} }$

${ k = \sqrt[3]{(16)^5 \cdot 9} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

${ k = \sqrt[3]{16^5 \cdot 3^2} }$

[ k = \sqrt[3]{16^5 \cdot

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In our previous article, we explored the solution to a logarithmic equation involving a base of $\frac{1}{k}$, where $k$ is a positive number greater than 1. In this article, we will provide a Q&A guide to help you better understand the solution and address any questions you may have.

Q: What is the main concept behind solving logarithmic equations?

A: The main concept behind solving logarithmic equations is to use the properties of logarithms to simplify the equation and isolate the variable.

Q: How do I simplify a logarithmic equation?

A: To simplify a logarithmic equation, you can use the properties of logarithms, such as the product rule, quotient rule, and power rule. You can also use algebraic manipulations, such as factoring and combining like terms.

Q: What is the difference of squares formula?

A: The difference of squares formula is a mathematical formula that states $(a + b)(a - b) = a^2 - b^2$. This formula can be used to simplify expressions involving the difference of two squares.

Q: How do I use the difference of squares formula to simplify an expression?

A: To use the difference of squares formula, you can let $a = 2\sqrt{2 + \sqrt{3}}$ and $b = 2\sqrt{2 - \sqrt{3}}$. Then, you can apply the formula to simplify the expression.

Q: What is the property of logarithms that states $\log_b a^c = c \log_b a$?

A: This property of logarithms states that the logarithm of a power is equal to the exponent multiplied by the logarithm of the base. In other words, $\log_b a^c = c \log_b a$.

Q: How do I use the property of logarithms to simplify an expression?

A: To use the property of logarithms, you can rewrite the expression as a product of logarithms. Then, you can use the property to simplify the expression.

Q: What is the cube root of a number?

A: The cube root of a number is a value that, when multiplied by itself three times, gives the original number. In other words, $\sqrt[3]{a} = b$ means that $b^3 = a$.

Q: How do I find the cube root of a number?

A: To find the cube root of a number, you can use a calculator or a mathematical formula. The cube root of a number can be found by raising the number to the power of $\frac{1}{3}$.

Q: What is the final solution to the logarithmic equation?

A: The final solution to the logarithmic equation is $k = \sqrt[3]{16^5 \cdot 3^2}$.

Q: What is the value of $k$?

A: The value of $k$ is a positive number greater than 1.

Q: Can I use this solution to solve other logarithmic equations?

A: Yes, you can use this solution as a guide to solve other logarithmic equations. However, you may need to modify the solution to fit the specific equation you are working with.

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 properties of logarithms to simplify the equation
• Not isolating the variable
• Not checking the domain of the logarithmic function
• Not using a calculator or mathematical formula to find the cube root of a number

Q: How can I practice solving logarithmic equations?

A: You can practice solving logarithmic equations by working through examples and exercises. You can also use online resources, such as calculators and mathematical software, to help you solve logarithmic equations.

Q: What are some real-world applications of logarithmic equations?

A: Logarithmic equations have many real-world applications, including:

• Finance: Logarithmic equations are used to calculate interest rates and investment returns.
• Science: Logarithmic equations are used to model population growth and decay.
• Engineering: Logarithmic equations are used to design and optimize systems.

Q: Can I use logarithmic equations to solve other types of equations?

A: Yes, you can use logarithmic equations to solve other types of equations, such as exponential equations and trigonometric equations. However, you may need to modify the solution to fit the specific equation you are working with.

Q: What are some common types of logarithmic equations?

A: Some common types of logarithmic equations include:

• Logarithmic equations with a base of 10
• Logarithmic equations with a base of e
• Logarithmic equations with a base of a variable

Q: How can I use logarithmic equations to model real-world phenomena?

A: You can use logarithmic equations to model real-world phenomena, such as population growth and decay, by using the properties of logarithms to simplify the equation and isolate the variable.

Q: What are some common mistakes to avoid when using logarithmic equations to model real-world phenomena?

A: Some common mistakes to avoid when using logarithmic equations to model real-world phenomena include:

• Not using the properties of logarithms to simplify the equation
• Not isolating the variable
• Not checking the domain of the logarithmic function
• Not using a calculator or mathematical formula to find the cube root of a number