Solve The Equation: $\log _5(x-3)=1-\log _5(x-7$\]If There Is More Than One Solution, Separate Them With Commas. If There Is No Solution, Select No Solution.$x=$No Solution

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Introduction

In this article, we will delve into solving a logarithmic equation involving base 5 logarithms. The equation is $\log _5(x-3)=1-\log _5(x-7)$. We will use various mathematical techniques to isolate the variable $x$ and find its possible values. This equation may have one or more solutions, or it may have no solution at all.

Understanding Logarithmic Equations

Before we proceed with solving the equation, let's briefly review what logarithmic equations are and how to work with them. A logarithmic equation is an equation that involves a logarithm, which is the inverse operation of exponentiation. In other words, if $y = \log_b(x)$, then $b^y = x$. The base of the logarithm is denoted by $b$, and the argument of the logarithm is denoted by $x$.

Step 1: Simplify the Equation

To simplify the equation, we can start by combining the logarithms on the right-hand side using the property $\log_b(x) - \log_b(y) = \log_b(\frac{x}{y})$. This gives us:

$$\log _5(x-3)=\log _5\left(\frac{x-7}{5}\right)$$

Step 2: Equate the Arguments

Since the logarithms have the same base, we can equate the arguments of the logarithms. This gives us:

$$x-3=\frac{x-7}{5}$$

Step 3: Solve for $x$

To solve for $x$, we can start by multiplying both sides of the equation by 5 to eliminate the fraction. This gives us:

$$5(x-3)=x-7$$

Expanding the left-hand side, we get:

$$5x-15=x-7$$

Subtracting $x$ from both sides, we get:

$$4x-15=-7$$

Adding 15 to both sides, we get:

$$4x=8$$

Dividing both sides by 4, we get:

$$x=2$$

Step 4: Check the Solutions

However, we need to check if the solutions we obtained are valid. We can do this by plugging the solutions back into the original equation.

For $x=2$, we have:

$$\log _5(2-3)=\log _5\left(\frac{2-7}{5}\right)$$

$$\log _5(-1)=\log _5\left(\frac{-5}{5}\right)$$

$$\log _5(-1)=\log _5(-1)$$

This is true, so $x=2$ is a valid solution.

However, we also need to check if the solutions are in the domain of the logarithmic functions. For $\log _5(x-3)$, we need $x-3>0$, which gives us $x>3$. For $\log _5(x-7)$, we need $x-7>0$, which gives us $x>7$.

Since $x=2$ is not greater than 7, it is not a valid solution.

Conclusion

In conclusion, we have found that the equation $\log _5(x-3)=1-\log _5(x-7)$ has no solution. This is because the solutions we obtained are not valid, and there are no other solutions that satisfy the equation.

Final Answer

The final answer is: $x=$No solution

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Introduction

In our previous article, we solved the equation $\log _5(x-3)=1-\log _5(x-7)$ and found that it has no solution. However, we received many questions from readers who were interested in learning more about logarithmic equations and how to solve them. In this article, we will answer some of the most frequently asked questions about solving logarithmic equations.

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, if $y = \log_b(x)$, then $b^y = x$. The base of the logarithm is denoted by $b$, and the argument of the logarithm is denoted by $x$.

Q: How do I simplify a logarithmic equation?

A: To simplify a logarithmic equation, you can start by combining the logarithms using the property $\log_b(x) - \log_b(y) = \log_b(\frac{x}{y})$. You can also use the property $\log_b(x) = \frac{\log_c(x)}{\log_c(b)}$ to change the base of the logarithm.

Q: How do I solve a logarithmic equation?

A: To solve a logarithmic equation, you can start by isolating the logarithm on one side of the equation. Then, you can use the properties of logarithms to simplify the equation and solve for the variable.

Q: What is the domain of a logarithmic function?

A: The domain of a logarithmic function is the set of all values of the variable that make the argument of the logarithm positive. For example, the domain of $\log_5(x-3)$ is $x>3$.

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

A: To check if a solution is valid, you can plug the solution back into the original equation and check if it is true. You can also check if the solution is in the domain of the logarithmic function.

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

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

• Not checking if the solution is in the domain of the logarithmic function
• Not plugging the solution back into the original equation to check if it is true
• Not using the properties of logarithms to simplify the equation
• Not being careful when working with fractions and decimals

Q: Can you give an example of a logarithmic equation that has a solution?

A: Yes, here is an example of a logarithmic equation that has a solution:

$$\log_2(x+1)=2$$

To solve this equation, we can start by isolating the logarithm on one side of the equation:

$$\log_2(x+1)=2$$

$$x+1=2^2$$

$$x+1=4$$

Subtracting 1 from both sides, we get:

$$x=3$$

This is a valid solution, since $x=3$ is in the domain of the logarithmic function.

Conclusion

In conclusion, solving logarithmic equations can be a challenging but rewarding task. By understanding the properties of logarithms and how to simplify and solve logarithmic equations, you can become proficient in solving these types of equations. Remember to always check if the solution is in the domain of the logarithmic function and to plug the solution back into the original equation to check if it is true.

Final Answer

The final answer is: There is no final numerical answer to this problem, as it is a Q&A article.