Consider The Equation $\log (3x-1) = \log_2 8$. Explain Why $3x-1$ Is Not Equal To 8. Describe The Steps You Would Take To Solve The Equation, And State What $3x-1$ Is Equal To.

Introduction

Logarithmic equations can be challenging to solve, but with a clear understanding of the properties of logarithms, we can break them down into manageable steps. In this article, we will explore the equation $\log (3x-1) = \log_2 8$ and explain why $3x-1$ is not equal to 8. We will then describe the steps to solve the equation and determine the value of $3x-1$.

Understanding the Equation

The given equation is $\log (3x-1) = \log_2 8$. At first glance, it may seem like a simple equation, but it requires careful analysis to solve. The key to solving this equation lies in understanding the properties of logarithms.

Why $3x-1$ is Not Equal to 8

Before we dive into solving the equation, let's address the question of why $3x-1$ is not equal to 8. This is a common misconception that can arise when dealing with logarithmic equations. The reason $3x-1$ is not equal to 8 is that the logarithm function is not one-to-one. In other words, there are multiple values of $x$ that can produce the same output for a given logarithmic expression.

To illustrate this, consider the equation $\log (3x-1) = \log 8$. If we were to assume that $3x-1 = 8$, we would be making an incorrect assumption. This is because the logarithm function is not injective (one-to-one), meaning that there are multiple values of $x$ that can produce the same output.

Solving the Equation

Now that we have addressed the misconception, let's move on to solving the equation. The first step is to apply the property of logarithms that states $\log_a b = \log_c b$ if and only if $a = c$. In this case, we can rewrite the equation as $\log (3x-1) = \log 8$.

Next, we can apply the property of logarithms that states $\log a = \log b$ if and only if $a = b$. This means that we can equate the expressions inside the logarithms, giving us $3x-1 = 8$.

However, as we discussed earlier, this is not the correct solution. Instead, we need to use the property of logarithms that states $\log a = \log b$ if and only if $a = b$. This means that we can equate the expressions inside the logarithms, giving us $3x-1 = 2^3$.

Simplifying the Equation

Now that we have simplified the equation, we can solve for $x$. The equation $3x-1 = 2^3$ can be rewritten as $3x-1 = 8$. To solve for $x$, we can add 1 to both sides of the equation, giving us $3x = 9$.

Next, we can divide both sides of the equation by 3, giving us $x = 3$.

Conclusion

In conclusion, the equation $\log (3x-1) = \log_2 8$ can be solved by applying the properties of logarithms. We found that $3x-1$ is not equal to 8, and instead, we solved for $x$ by equating the expressions inside the logarithms. The final solution is $x = 3$, and $3x-1 = 8$.

Properties of Logarithms

The properties of logarithms are essential for solving logarithmic equations. Here are some key properties to keep in mind:

• Logarithm of a product: $\log (ab) = \log a + \log b$
• Logarithm of a quotient: $\log \left(\frac{a}{b}\right) = \log a - \log b$
• Logarithm of a power: $\log a^b = b \log a$
• Change of base: $\log_a b = \frac{\log_c b}{\log_c a}$

Common Logarithms

Common logarithms are logarithms with a base of 10. They are denoted by $\log$ without a subscript. For example, $\log 10 = 1$ and $\log 100 = 2$.

Natural Logarithms

Natural logarithms are logarithms with a base of $e$. They are denoted by $\ln$. For example, $\ln e = 1$ and $\ln e^2 = 2$.

Solving Logarithmic Equations

Solving logarithmic equations requires a clear understanding of the properties of logarithms. Here are some steps to follow:

1. Apply the properties of logarithms: Use the properties of logarithms to simplify the equation.
2. Equating expressions: Equate the expressions inside the logarithms.
3. Solving for x: Solve for $x$ by isolating it on one side of the equation.

Example Problems

Here are some example problems to help you practice solving logarithmic equations:

• $\log (2x+1) = \log 16$
• $\log_3 (x-2) = \log_3 9$
• $\log (x+1) = \log 25$

Conclusion

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

A: A logarithmic equation is an equation that involves a logarithmic expression, while an exponential equation is an equation that involves an exponential expression. For example, $\log (3x-1) = \log_2 8$ is a logarithmic equation, while $2^x = 8$ 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: $\log_a b = \frac{\log_c b}{\log_c a}$. For example, to solve $\log_2 (3x-1) = \log_2 8$, you can use the change of base formula to rewrite the equation as $\log (3x-1) = \log 8$.

Q: Can I use the properties of logarithms to simplify a logarithmic equation?

A: Yes, you can use the properties of logarithms to simplify a logarithmic equation. For example, if you have the equation $\log (2x+1) + \log (3x-1) = \log 16$, you can use the property of logarithms that states $\log a + \log b = \log ab$ to simplify the equation to $\log (2x+1)(3x-1) = \log 16$.

Q: How do I solve a logarithmic equation with a logarithm of a power?

A: To solve a logarithmic equation with a logarithm of a power, you can use the property of logarithms that states $\log a^b = b \log a$. For example, to solve $\log (3x-1)^2 = \log 64$, you can use the property of logarithms to rewrite the equation as $2 \log (3x-1) = \log 64$.

Q: Can I use the properties of logarithms to solve a logarithmic equation with a logarithm of a quotient?

A: Yes, you can use the properties of logarithms to solve a logarithmic equation with a logarithm of a quotient. For example, if you have the equation $\log \left(\frac{2x+1}{3x-1}\right) = \log 4$, you can use the property of logarithms that states $\log \frac{a}{b} = \log a - \log b$ to simplify the equation to $\log (2x+1) - \log (3x-1) = \log 4$.

Q: How do I solve a logarithmic equation with a logarithm of a product?

A: To solve a logarithmic equation with a logarithm of a product, you can use the property of logarithms that states $\log ab = \log a + \log b$. For example, to solve $\log (2x+1)(3x-1) = \log 16$, you can use the property of logarithms to rewrite the equation as $\log (2x+1) + \log (3x-1) = \log 16$.

Q: Can I use the properties of logarithms to solve a logarithmic equation with a logarithm of a power and a logarithm of a quotient?

A: Yes, you can use the properties of logarithms to solve a logarithmic equation with a logarithm of a power and a logarithm of a quotient. For example, if you have the equation $\log (2x+1)^2 - \log (3x-1) = \log 4$, you can use the properties of logarithms to simplify the equation to $2 \log (2x+1) - \log (3x-1) = \log 4$.

Q: How do I know which property of logarithms to use when solving a logarithmic equation?

A: To determine which property of logarithms to use when solving a logarithmic equation, you need to analyze the equation and identify the type of logarithmic expression involved. For example, if the equation involves a logarithm of a product, you can use the property of logarithms that states $\log ab = \log a + \log b$. If the equation involves a logarithm of a quotient, you can use the property of logarithms that states $\log \frac{a}{b} = \log a - \log b$.

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 essential to understand the properties of logarithms and how to apply them to solve the equation. Using a calculator can help you verify your solution and ensure that you have the correct answer.

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

A: To check your solution to a logarithmic equation, you can substitute the value of $x$ back into the original equation and verify that it is true. For example, if you have the equation $\log (3x-1) = \log_2 8$ and you find that $x = 3$, you can substitute $x = 3$ back into the original equation to verify that it is true.