Solving the Logarithmic Equation

Oct 13, 2025 by recoveryshake.com 123 views

Introduction

Logarithmic equations can be challenging to solve, especially when dealing with extraneous solutions. In this article, we will explore the concept of extraneous solutions in logarithmic equations and provide a step-by-step guide on how to identify them. We will also apply this knowledge to solve the given logarithmic equation and determine the correct answer.

What are Extraneous Solutions?

Extraneous solutions are values that satisfy an equation but are not valid solutions. In the context of logarithmic equations, extraneous solutions occur when the argument of the logarithm is negative or zero. This is because the logarithm of a negative or zero number is undefined.

Properties of Logarithms

Before we dive into solving the given equation, it's essential to recall some properties of logarithms. The two main properties we will use are:

Product Property: $\log_b(MN) = \log_b(M) + \log_b(N)$
Quotient Property: $\log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)$

Solving the Logarithmic Equation

Now that we have a basic understanding of logarithmic properties and extraneous solutions, let's tackle the given equation:

$\log_7(3x^3 + x) - \log_7(x) = 2$

Our goal is to solve for $x$ . To do this, we can use the quotient property to combine the two logarithms on the left-hand side:

$\log_7(3x^3 + x) - \log_7(x) = \log_7\left(\frac{3x^3 + x}{x}\right) = 2$

Next, we can rewrite the equation in exponential form:

$7^2 = \frac{3x^3 + x}{x}$

Simplifying the equation, we get:

$49 = 3x^2 + 1$

Subtracting 1 from both sides gives us:

$48 = 3x^2$

Dividing both sides by 3, we get:

$16 = x^2$

Taking the square root of both sides, we get:

$x = \pm 4$

Identifying Extraneous Solutions

Now that we have found the solutions to the equation, we need to determine which one is an extraneous solution. To do this, we can substitute each solution back into the original equation and check if it satisfies the equation.

Let's start with $x = 4$ . Substituting this value into the original equation, we get:

$\log_7(3(4)^3 + 4) - \log_7(4) = \log_7(3(64) + 4) - \log_7(4)$

Simplifying the equation, we get:

$\log_7(192 + 4) - \log_7(4) = \log_7(196) - \log_7(4)$

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

$\log_7(196) - \log_7(4) = \log_7\left(\frac{196}{4}\right)$

Simplifying the equation, we get:

$\log_7(49) = \log_7(49)$

This equation is true, so $x = 4$ is a valid solution.

Now, let's try $x = -4$ . Substituting this value into the original equation, we get:

$\log_7(3(-4)^3 + (-4)) - \log_7(-4) = \log_7(3(-64) - 4) - \log_7(-4)$

Simplifying the equation, we get:

$\log_7(-192 - 4) - \log_7(-4) = \log_7(-196) - \log_7(-4)$

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

$\log_7(-196) - \log_7(-4) = \log_7\left(\frac{-196}{-4}\right)$

Simplifying the equation, we get:

$\log_7(49) = \log_7(49)$

This equation is true, so $x = -4$ is also a valid solution.

However, we need to check if $x = -4$ is an extraneous solution. To do this, we can substitute $x = -4$ back into the original equation and check if the argument of the logarithm is negative or zero.

Substituting $x = -4$ into the original equation, we get:

$\log_7(3(-4)^3 + (-4)) = \log_7(3(-64) - 4)$

Simplifying the equation, we get:

$\log_7(-192 - 4) = \log_7(-196)$

The argument of the logarithm is negative, so $x = -4$ is an extraneous solution.

Conclusion

In this article, we explored the concept of extraneous solutions in logarithmic equations and provided a step-by-step guide on how to identify them. We applied this knowledge to solve the given logarithmic equation and determined that $x = -4$ is an extraneous solution.

Answer

The correct answer is:

B. $x = -4$

Q: What is an extraneous solution in a logarithmic equation?

A: An extraneous solution is a value that satisfies a logarithmic equation but is not a valid solution. This occurs when the argument of the logarithm is negative or zero.

Q: How do I identify an extraneous solution in a logarithmic equation?

A: To identify an extraneous solution, substitute the solution back into the original equation and check if the argument of the logarithm is negative or zero. If it is, then the solution is an extraneous solution.

Q: What are some common properties of logarithms that I should know?

A: Some common properties of logarithms include:

Product Property: $\log_b(MN) = \log_b(M) + \log_b(N)$
Quotient Property: $\log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)$
Power Property: $\log_b(M^p) = p\log_b(M)$

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, use the change of base formula:

$\log_b(M) = \frac{\log_c(M)}{\log_c(b)}$

where $c$ is any positive real number.

Q: What is the change of base formula?

A: The change of base formula is:

$\log_b(M) = \frac{\log_c(M)}{\log_c(b)}$

where $c$ is any positive real number.

Q: How do I use the change of base formula to solve a logarithmic equation?

A: To use the change of base formula, substitute the change of base formula into the original equation and simplify.

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 for extraneous solutions
Not using the correct properties of logarithms
Not simplifying the equation correctly

Q: How do I check for extraneous solutions in a logarithmic equation?

A: To check for extraneous solutions, substitute the solution back into the original equation and check if the argument of the logarithm is negative or zero. If it is, then the solution is an extraneous solution.

Q: What is the final answer to the original problem?

A: The final answer to the original problem is:

B. $x = -4$

Note that $x = 4$ is a valid solution, but $x = -4$ is an extraneous solution.