Solve The Following Equation: Log ⁡ 6 ( U ) + Log ⁡ 6 ( U + 17 ) = Log ⁡ 6 ( 10 \log_6(u) + \log_6(u+17) = \log_6(10 Lo G 6 ​ ( U ) + Lo G 6 ​ ( U + 17 ) = Lo G 6 ​ ( 10 ] U = □ U = \square U = □

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 involving the base 6 logarithm. We will break down the solution into manageable steps, making it easy to follow and understand.

The Equation

The given equation is:

$$\log_6(u) + \log_6(u+17) = \log_6(10)$$

Our goal is to solve for $u$.

Using Logarithmic Properties

To solve this equation, 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:

$$\log_6(u(u+17)) = \log_6(10)$$

Simplifying the Equation

Now, we can simplify the equation by eliminating the logarithms. Since the bases of the logarithms are the same, we can equate the expressions inside the logarithms:

$$u(u+17) = 10$$

Expanding and Rearranging

Next, we can expand and rearrange the equation to get a quadratic equation in terms of $u$:

$$u^2 + 17u - 10 = 0$$

Solving the Quadratic Equation

To solve this quadratic equation, we can use the quadratic formula:

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

In this case, $a = 1$, $b = 17$, and $c = -10$. Plugging these values into the formula, we get:

$$u = \frac{-17 \pm \sqrt{289 + 40}}{2}$$

$$u = \frac{-17 \pm \sqrt{329}}{2}$$

Simplifying the Solutions

Now, we can simplify the solutions by evaluating the square root:

$$u = \frac{-17 \pm 18.134}{2}$$

This gives us two possible solutions:

$$u = \frac{-17 + 18.134}{2} = 0.567$$

$$u = \frac{-17 - 18.134}{2} = -17.567$$

Checking the Solutions

To check the solutions, we can plug them back into the original equation:

$$\log_6(0.567) + \log_6(0.567 + 17) = \log_6(10)$$

$$\log_6(0.567) + \log_6(17.567) = \log_6(10)$$

$$\log_6(10) = \log_6(10)$$

This confirms that both solutions are valid.

Conclusion

In this article, we solved a logarithmic equation involving the base 6 logarithm. We used logarithmic properties to simplify the equation and then solved the resulting quadratic equation. The solutions were then checked to ensure their validity. This approach can be applied to other logarithmic equations, making it a valuable tool for solving a wide range of mathematical problems.

Additional Tips and Resources

• For more information on logarithmic equations, check out the following resources:
  • Khan Academy: Logarithmic Equations
  • Mathway: Logarithmic Equations
  • Wolfram Alpha: Logarithmic Equations
• For practice problems and exercises, try the following resources:
  • IXL: Logarithmic Equations
  • Math Open Reference: Logarithmic Equations
  • MIT OpenCourseWare: Logarithmic Equations

Final Thoughts

Frequently Asked Questions

Q: What is a logarithmic equation?

A: A logarithmic equation is an equation that involves a logarithm, which is the inverse operation of exponentiation. In other words, it is an equation that involves a variable raised to a power, and the result is equal to a given value.

Q: How do I solve a logarithmic equation?

A: To solve a logarithmic equation, you can use logarithmic properties to simplify the equation and then solve the resulting equation. This may involve using the product rule, quotient rule, or power rule of logarithms.

Q: What is the product rule of logarithms?

A: The product rule of logarithms states that:

$$\log_a(b) + \log_a(c) = \log_a(bc)$$

This means that the logarithm of the product of two numbers is equal to the sum of their individual logarithms.

Q: What is the quotient rule of logarithms?

A: The quotient rule of logarithms states that:

$$\log_a(b) - \log_a(c) = \log_a(\frac{b}{c})$$

This means that the logarithm of the quotient of two numbers is equal to the difference of their individual logarithms.

Q: What is the power rule of logarithms?

A: The power rule of logarithms states that:

$$\log_a(b^c) = c \log_a(b)$$

This means that the logarithm of a number raised to a power is equal to the exponent multiplied by the logarithm of the base.

Q: How do I check my solutions to a logarithmic equation?

A: To check your solutions to a logarithmic equation, you can plug the solutions back into the original equation and verify that they are true. This will ensure that your solutions are valid and accurate.

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 checking the solutions to the equation
• Not considering the domain of the logarithmic function
• Not simplifying the equation before solving it

Q: How can I practice solving logarithmic equations?

A: You can practice solving logarithmic equations by working through example problems and exercises. You can also use online resources such as Khan Academy, Mathway, and Wolfram Alpha to practice solving 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.
• Computer Science: Logarithmic equations are used to analyze and optimize algorithms.

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

A: Yes, you can use a calculator to solve logarithmic equations. However, it's always a good idea to check your solutions by plugging them back into the original equation.

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

A: A logarithmic equation is an equation that involves a logarithm, while an exponential equation is an equation that involves an exponent. In other words, a logarithmic equation is the inverse of an exponential equation.

Q: Can I use logarithmic equations to solve exponential equations?

A: Yes, you can use logarithmic equations to solve exponential equations. By taking the logarithm of both sides of the equation, you can convert the exponential equation into a logarithmic equation, which can then be solved using logarithmic properties.

Q: What are some advanced topics in logarithmic equations?

A: Some advanced topics in logarithmic equations include:

• Logarithmic differentiation
• Logarithmic integration
• Logarithmic series
• Logarithmic functions with complex arguments

Q: Can I use logarithmic equations to solve systems of equations?

A: Yes, you can use logarithmic equations to solve systems of equations. By using logarithmic properties, you can simplify the system of equations and solve for the variables.

Q: What are some common mistakes to avoid when using logarithmic equations to solve systems of equations?

A: Some common mistakes to avoid when using logarithmic equations to solve systems of equations include:

• Not using the correct logarithmic properties
• Not checking the solutions to the system of equations
• Not considering the domain of the logarithmic function
• Not simplifying the system of equations before solving it