Solving Logarithmic Equations: A Step-by-Step Guide

Hey there, math enthusiasts! Today, we're diving into the world of logarithmic equations with a fun and engaging approach. Specifically, we're tackling the equation $\log _{x+7}(3x^2-11)=2$. Don't worry if it looks intimidating; we'll break it down into manageable steps. By the end of this guide, you'll not only know how to solve this particular problem but also gain a solid understanding of the underlying principles. So, grab your favorite beverage, settle in, and let's get started!

Understanding the Basics of Logarithms

Before we jump into solving the equation, let's quickly review what logarithms are all about. A logarithm is essentially the inverse operation of exponentiation. In simpler terms, if we have an equation like $a^b = c$, then we can express this relationship using logarithms as $\log_a c = b$. Here, 'a' is the base, 'b' is the exponent, and 'c' is the result.

Why are logarithms useful? Well, they help us solve for exponents in equations. They're also incredibly handy in various fields, including science, engineering, and finance. Understanding logarithms opens doors to solving complex problems and analyzing data more effectively.

In our equation, $\log _{x+7}(3x^2-11)=2$, we need to remember that $(x+7)$ is the base of the logarithm, and $(3x^2-11)$ is the argument. The result of the logarithm is 2. Our goal is to find the value(s) of 'x' that make this equation true. Keep in mind that the base of a logarithm must be positive and not equal to 1, and the argument must be positive. These are crucial conditions that we'll need to check later to ensure our solutions are valid.

Logarithmic functions are used to measure many things around us. For example, the richter scale is a base 10 logarithmic scale. Decibels are another use of log scales. These applications are useful because they can reduce exponential growth to a smaller scale that is easier to interpret. Understanding these real-world applications can help you solidify your understanding of logarithmic functions and their relevance.

Converting the Logarithmic Equation to Exponential Form

The first step in solving our equation is to convert it from logarithmic form to exponential form. Remember the relationship $a^b = c$ and $\log_a c = b$? We can apply this directly to our equation, $\log _{x+7}(3x^2-11)=2$.

Here, $a = (x+7)$, $b = 2$, and $c = (3x^2-11)$. So, we can rewrite the equation as:

$(x+7)^2 = 3x^2 - 11$

Now, we have a quadratic equation that we can solve. This conversion is a critical step because it transforms the logarithmic equation into a more familiar algebraic form. From here, we can use standard algebraic techniques to find the value(s) of 'x'. Always double-check your conversion to ensure accuracy, as a mistake here can lead to incorrect solutions.

Expanding the left side, we get $(x+7)(x+7) = x^2 + 14x + 49$. So now the equation becomes:

$x^2 + 14x + 49 = 3x^2 - 11$

Solving the Quadratic Equation

Now that we have the equation $x^2 + 14x + 49 = 3x^2 - 11$, we need to rearrange it into the standard quadratic form, which is $ax^2 + bx + c = 0$. To do this, let's move all the terms to one side of the equation. Subtracting $x^2$, $14x$, and $49$ from both sides, we get:

$0 = 2x^2 - 14x - 60$

Now, we can simplify this equation by dividing all the terms by 2:

$0 = x^2 - 7x - 30$

We can solve this quadratic equation by factoring, using the quadratic formula, or completing the square. Factoring is often the quickest method if we can find two numbers that multiply to -30 and add up to -7. In this case, those numbers are -10 and 3. So, we can factor the equation as:

$0 = (x - 10)(x + 3)$

Setting each factor equal to zero gives us two possible solutions for 'x':

$x - 10 = 0 \Rightarrow x = 10$

$x + 3 = 0 \Rightarrow x = -3$

So, our potential solutions are $x = 10$ and $x = -3$. But remember, we need to check these solutions to make sure they are valid in the original logarithmic equation.

Remember to always rearrange the equation so that one side equals 0. This allows for factoring or utilizing the quadratic formula effectively. Mastering these algebraic manipulations is crucial for solving quadratic equations accurately.

Checking for Extraneous Solutions

We found two potential solutions: $x = 10$ and $x = -3$. However, not all solutions we find in the process of solving an equation are necessarily valid. These are called extraneous solutions. In the context of logarithmic equations, extraneous solutions often arise because the domain of a logarithmic function is restricted. Specifically, the base of the logarithm must be positive and not equal to 1, and the argument must be positive.

Let's check our solutions against these conditions:

1. For $x = 10$:

• Base: $x + 7 = 10 + 7 = 17$. This is positive and not equal to 1, so it's valid.
• Argument: $3x^2 - 11 = 3(10)^2 - 11 = 300 - 11 = 289$. This is positive, so it's also valid.

Therefore, $x = 10$ is a valid solution.

2. For $x = -3$:

• Base: $x + 7 = -3 + 7 = 4$. This is positive and not equal to 1, so it's valid.
• Argument: $3x^2 - 11 = 3(-3)^2 - 11 = 3(9) - 11 = 27 - 11 = 16$. This is positive, so it's also valid.

Therefore, $x = -3$ is also a valid solution.

Always remember to check your solutions! This step is often overlooked but is crucial to ensure the validity of your answers. Substituting the solutions back into the original equation and verifying the conditions for the base and argument of the logarithm will help you avoid extraneous solutions.

Final Solution

After carefully checking our potential solutions, we've determined that both $x = 10$ and $x = -3$ are valid solutions to the equation $\log _{x+7}(3x^2-11)=2$.

Therefore, the final solution set is:

$x = \{-3, 10\}$

Congratulations! You've successfully solved a logarithmic equation. Remember, the key is to understand the relationship between logarithms and exponents, convert the equation to a more manageable form, solve for the variable, and always check for extraneous solutions.

By mastering these steps, you'll be well-equipped to tackle a wide range of logarithmic equations. Keep practicing, and you'll become more confident in your ability to solve these types of problems.

In conclusion, solving logarithmic equations involves a systematic approach. Start by understanding the basics of logarithms, convert the equation to exponential form, solve the resulting algebraic equation, and, most importantly, check for extraneous solutions. With practice and a solid understanding of the underlying principles, you'll be able to solve logarithmic equations with ease.

For further reading and to deepen your understanding of logarithmic functions, consider visiting Khan Academy's Logarithm Section. This resource provides comprehensive lessons, practice exercises, and videos to help you master logarithms.