Solve The Inequality: $\[\log _3\left(x^2-3x-67\right)\ \textless \ 3\\]

$$

Introduction

In this article, we will delve into the world of logarithmic inequalities and explore the solution to the given inequality: $\log_3(x^2-3x-67) < 3$. This type of inequality involves logarithmic functions and requires a deep understanding of the properties of logarithms. We will break down the solution step by step, using algebraic manipulations and properties of logarithms to arrive at the final solution.

Understanding the Inequality

The given inequality is $\log_3(x^2-3x-67) < 3$. To begin solving this inequality, we need to understand the properties of logarithmic functions. The logarithmic function $\log_b(x)$ is defined as the exponent to which the base $b$ must be raised to produce the number $x$. In this case, the base is $3$, and the argument of the logarithm is $x^2-3x-67$.

Step 1: Exponentiating Both Sides

To solve the inequality, we can start by exponentiating both sides of the inequality. This will allow us to eliminate the logarithm and work with a polynomial inequality. We have:

$$\log_3(x^2-3x-67) < 3$$

Exponentiating both sides with base $3$, we get:

$$x^2-3x-67 < 3^3$$

Step 2: Simplifying the Right-Hand Side

The right-hand side of the inequality is $3^3$, which is equal to $27$. Therefore, we can simplify the inequality as follows:

$$x^2-3x-67 < 27$$

Step 3: Rearranging the Inequality

To make it easier to work with, we can rearrange the inequality by subtracting $27$ from both sides:

$$x^2-3x-94 < 0$$

Step 4: Factoring the Quadratic

The left-hand side of the inequality is a quadratic expression that can be factored. We have:

$$x^2-3x-94 = (x-14)(x+7)$$

Step 5: Finding the Critical Points

To find the critical points of the inequality, we need to set the factors equal to zero and solve for $x$. We have:

$$(x-14)(x+7) = 0$$

Solving for $x$, we get:

$$x-14 = 0 \quad \text{or} \quad x+7 = 0$$

$$x = 14 \quad \text{or} \quad x = -7$$

Step 6: Testing the Intervals

To determine the solution to the inequality, we need to test the intervals between the critical points. We have three intervals to test: $(-\infty, -7)$, $(-7, 14)$, and $(14, \infty)$.

Step 7: Testing the Interval $(-\infty, -7)$

To test the interval $(-\infty, -7)$, we can choose a test value, say $x = -10$. Plugging this value into the inequality, we get:

$$(x-14)(x+7) = (-10-14)(-10+7) = (-24)(-3) = 72$$

Since $72 > 0$, the inequality is true for this interval.

Step 8: Testing the Interval $(-7, 14)$

To test the interval $(-7, 14)$, we can choose a test value, say $x = 0$. Plugging this value into the inequality, we get:

$$(x-14)(x+7) = (0-14)(0+7) = (-14)(7) = -98$$

Since $-98 < 0$, the inequality is false for this interval.

Step 9: Testing the Interval $(14, \infty)$

To test the interval $(14, \infty)$, we can choose a test value, say $x = 20$. Plugging this value into the inequality, we get:

$$(x-14)(x+7) = (20-14)(20+7) = (6)(27) = 162$$

Since $162 > 0$, the inequality is true for this interval.

Conclusion

In conclusion, the solution to the inequality $\log_3(x^2-3x-67) < 3$ is $x \in (-\infty, -7) \cup (14, \infty)$. This means that the inequality is true for all values of $x$ less than $-7$ and all values of $x$ greater than $14$.

Final Answer

The final answer is $\boxed{(-\infty, -7) \cup (14, \infty)}$.

$$

Introduction

In our previous article, we solved the inequality $\log_3(x^2-3x-67) < 3$ and found that the solution is $x \in (-\infty, -7) \cup (14, \infty)$. In this article, we will answer some frequently asked questions related to the solution of this inequality.

Q&A

Q: What is the meaning of the inequality $\log_3(x^2-3x-67) < 3$?

A: The inequality $\log_3(x^2-3x-67) < 3$ means that the logarithm of the expression $x^2-3x-67$ with base $3$ is less than $3$. In other words, the expression $x^2-3x-67$ must be less than $3^3$, which is equal to $27$.

Q: How did you solve the inequality?

A: We solved the inequality by exponentiating both sides with base $3$, which eliminated the logarithm and allowed us to work with a polynomial inequality. We then simplified the inequality, factored the quadratic expression, and found the critical points.

Q: What are the critical points of the inequality?

A: The critical points of the inequality are the values of $x$ that make the factors of the quadratic expression equal to zero. In this case, the critical points are $x = 14$ and $x = -7$.

Q: How do you determine the solution to the inequality?

A: To determine the solution to the inequality, we need to test the intervals between the critical points. We choose a test value from each interval and plug it into the inequality to see if it is true or false.

Q: What is the solution to the inequality?

A: The solution to the inequality is $x \in (-\infty, -7) \cup (14, \infty)$. This means that the inequality is true for all values of $x$ less than $-7$ and all values of $x$ greater than $14$.

Q: Why is the inequality true for $x < -7$?

A: The inequality is true for $x < -7$ because the expression $x^2-3x-67$ is always negative for $x < -7$. When we take the logarithm of a negative number, we get a complex number, which is less than $3$.

Q: Why is the inequality true for $x > 14$?

A: The inequality is true for $x > 14$ because the expression $x^2-3x-67$ is always positive for $x > 14$. When we take the logarithm of a positive number, we get a real number, which is less than $3$.

Q: What is the significance of the inequality $\log_3(x^2-3x-67) < 3$?

A: The inequality $\log_3(x^2-3x-67) < 3$ has significant implications in various fields, including mathematics, physics, and engineering. It can be used to model real-world problems, such as population growth, chemical reactions, and electrical circuits.

Conclusion

In conclusion, the inequality $\log_3(x^2-3x-67) < 3$ has a solution of $x \in (-\infty, -7) \cup (14, \infty)$. We answered some frequently asked questions related to the solution of this inequality and provided a detailed explanation of the steps involved in solving it.

Final Answer

The final answer is $\boxed{(-\infty, -7) \cup (14, \infty)}$.