Solve For X X X : Log ⁡ ( X ) + Log ⁡ ( X + 3 ) = 1 \log (x) + \log (x+3) = 1 Lo G ( X ) + Lo G ( X + 3 ) = 1 X = X = X =

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: $\log (x) + \log (x+3) = 1$. We will break down the solution into manageable steps, making it easy to understand and follow.

Understanding Logarithmic Properties

Before we dive into solving the equation, it's essential to understand the properties of logarithms. The logarithm of a number is the exponent to which a base number must be raised to produce that number. In this case, we are dealing with base 10 logarithms, which are the most common type of logarithm.

One of the key properties of logarithms is the product rule, which states that $\log (a) + \log (b) = \log (ab)$. This property will be crucial in solving our equation.

Applying the Product Rule

Using the product rule, we can rewrite the equation as:

$$\log (x) + \log (x+3) = \log (x(x+3))$$

This simplifies the equation, making it easier to work with.

Simplifying the Equation

Now that we have applied the product rule, we can simplify the equation further. We know that $\log (x(x+3)) = \log (x^2 + 3x)$.

So, our equation becomes:

$$\log (x^2 + 3x) = 1$$

Exponentiating Both Sides

To get rid of the logarithm, we can exponentiate both sides of the equation. Since we are dealing with base 10 logarithms, we can use the fact that $10^{\log (x)} = x$.

Exponentiating both sides, we get:

$$x^2 + 3x = 10^1$$

Simplifying the Right-Hand Side

The right-hand side of the equation is $10^1$, which is equal to 10.

So, our equation becomes:

$$x^2 + 3x = 10$$

Rearranging the Equation

To make it easier to solve, we can rearrange the equation by subtracting 10 from both sides:

$$x^2 + 3x - 10 = 0$$

Factoring the Quadratic

The quadratic equation $x^2 + 3x - 10 = 0$ can be factored as:

$$(x + 5)(x - 2) = 0$$

Solving for $x$

To solve for $x$, we can set each factor equal to zero and solve for $x$:

$$x + 5 = 0 \Rightarrow x = -5$$

$$x - 2 = 0 \Rightarrow x = 2$$

Conclusion

In this article, we have solved the logarithmic equation $\log (x) + \log (x+3) = 1$. We have applied the product rule, simplified the equation, exponentiated both sides, and finally solved for $x$ using factoring.

The two possible solutions for $x$ are $x = -5$ and $x = 2$. However, we need to check if these solutions are valid by plugging them back into the original equation.

Checking the Solutions

Let's plug $x = -5$ back into the original equation:

$$\log (-5) + \log (-5+3) = \log (-5) + \log (-2)$$

Since the logarithm of a negative number is undefined, $x = -5$ is not a valid solution.

Now, let's plug $x = 2$ back into the original equation:

$$\log (2) + \log (2+3) = \log (2) + \log (5)$$

Using the product rule, we can rewrite this as:

$$\log (2 \cdot 5) = \log (10)$$

Since $\log (10) = 1$, $x = 2$ is a valid solution.

Final Answer

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Introduction

In our previous article, we solved the logarithmic equation $\log (x) + \log (x+3) = 1$. We applied the product rule, simplified the equation, exponentiated both sides, and finally solved for $x$ using factoring.

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 product rule of logarithms?

A: The product rule of logarithms states that $\log (a) + \log (b) = \log (ab)$. This means that the logarithm of the product of two numbers is equal to the sum of their individual logarithms.

Q: How do I apply the product rule to a logarithmic equation?

A: To apply the product rule, simply rewrite the equation using the product rule formula: $\log (a) + \log (b) = \log (ab)$. Then, simplify the equation by combining the logarithms.

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 exponential function. For example, $\log (x) = 2$ is a logarithmic equation, while $x^2 = 4$ is an exponential equation.

Q: How do I solve a logarithmic equation with a base other than 10?

A: To solve a logarithmic equation with a base other than 10, you can use the change of base formula: $\log_b (x) = \frac{\log_a (x)}{\log_a (b)}$. This formula allows you to convert a logarithm with a base other than 10 to a logarithm with base 10.

Q: What is the relationship between logarithms and exponents?

A: Logarithms and exponents are inverse functions. This means that if $y = \log (x)$, then $x = 10^y$. Similarly, if $y = a^x$, then $x = \log_a (y)$.

Q: How do I check if a solution to a logarithmic equation is valid?

A: To check if a solution to a logarithmic equation is valid, plug the solution back into the original equation and simplify. If the solution satisfies the equation, then it is a valid solution.

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 apply the product rule
• Not simplifying the equation properly
• Not checking if the solution is valid
• Not using the correct base for the logarithm

Conclusion

In this Q&A guide, we have addressed some common questions and concerns about solving logarithmic equations. We hope that this guide has been helpful in clarifying any confusion and providing a better understanding of logarithmic equations.

Additional Resources

For more information on logarithmic equations, we recommend checking out the following resources:

• Khan Academy: Logarithmic Equations
• Mathway: Logarithmic Equations
• Wolfram Alpha: Logarithmic Equations

Final Tips

• Always apply the product rule when simplifying logarithmic equations
• Check if the solution is valid by plugging it back into the original equation
• Use the correct base for the logarithm
• Simplify the equation properly to avoid mistakes

By following these tips and using the resources provided, you will be well on your way to becoming proficient in solving logarithmic equations.