Solving For Equations Of Perpendicular Lines And Logarithmic Equations

Introduction

This article delves into two fundamental concepts in mathematics: finding the equation of a line perpendicular to a given line and solving systems of logarithmic equations. These topics are crucial for a solid foundation in algebra and calculus, and they often appear in various applications across science and engineering. We will explore the underlying principles, step-by-step solutions, and practical insights to help you master these skills. Understanding these concepts will not only improve your problem-solving abilities but also enhance your overall mathematical acumen. Let's embark on this journey to unravel the intricacies of perpendicular lines and logarithmic equations.

1. Finding the Equation of a Perpendicular Line

Perpendicular lines play a significant role in geometry and coordinate geometry. When two lines are perpendicular, their slopes have a special relationship. Specifically, the product of their slopes is -1. This property is crucial when we need to find the equation of a line that is perpendicular to another given line and passes through a specific point. Let's tackle the problem at hand: find the equation of the line which passes through the point $(3, -2)$ and is perpendicular to the line $3x + 2y - 4 = 0$.

To begin, we need to determine the slope of the given line. The equation $3x + 2y - 4 = 0$ is in the general form of a linear equation. To find the slope, we can rewrite this equation in the slope-intercept form, which is $y = mx + b$, where $m$ represents the slope and $b$ represents the y-intercept. Rearranging the given equation, we have:

$2y = -3x + 4$

Dividing both sides by 2, we get:

$y = -\frac{3}{2}x + 2$

From this, we can see that the slope of the given line is $-\frac{3}{2}$. Now, to find the slope of the line perpendicular to this, we need to take the negative reciprocal of $-\frac{3}{2}$. The negative reciprocal is found by flipping the fraction and changing its sign. Thus, the slope of the perpendicular line, which we'll call $m_{\perp}$, is:

$m_{\perp} = \frac{2}{3}$

Now that we have the slope of the perpendicular line, we can use the point-slope form of a linear equation to find its equation. The point-slope form is given by:

$y - y_1 = m(x - x_1)$

where $(x_1, y_1)$ is a point on the line and $m$ is the slope. We know that the perpendicular line passes through the point $(3, -2)$, so we can substitute $x_1 = 3$, $y_1 = -2$, and $m = \frac{2}{3}$ into the point-slope form:

$y - (-2) = \frac{2}{3}(x - 3)$

Simplifying, we get:

$y + 2 = \frac{2}{3}x - 2$

To write this equation in the slope-intercept form, we subtract 2 from both sides:

$y = \frac{2}{3}x - 4$

Alternatively, we can write the equation in the general form $Ax + By + C = 0$. Multiplying both sides by 3 to eliminate the fraction, we get:

$3y = 2x - 12$

Rearranging the terms, we have:

$2x - 3y - 12 = 0$

Thus, the equation of the line that passes through the point $(3, -2)$ and is perpendicular to the line $3x + 2y - 4 = 0$ is $y = \frac{2}{3}x - 4$ in slope-intercept form or $2x - 3y - 12 = 0$ in general form. This process exemplifies how understanding the relationship between slopes of perpendicular lines and utilizing the point-slope form can lead to the solution. The ability to manipulate linear equations and apply geometric principles is a cornerstone of mathematical problem-solving.

2. Solving Systems of Logarithmic Equations

Logarithmic equations are an integral part of algebra and calculus, often used in modeling phenomena in physics, engineering, and finance. Solving systems of logarithmic equations requires a good grasp of logarithmic properties and algebraic manipulation. In this section, we will address the problem of solving for $x$ and $y$ in the following system of equations:

$\begin{cases} \log (x-1) + 2 \log y = 2 \log 3 \\ \log x + \log y = \log 5 \end{cases}$

To solve this system, we will use the properties of logarithms to simplify the equations and then employ algebraic techniques to find the values of $x$ and $y$. First, let's rewrite the equations using logarithmic properties. Recall that $a \log b = \log b^a$ and $\log a + \log b = \log (ab)$. Applying these properties, the first equation becomes:

$\log (x - 1) + \log y^2 = \log 3^2$

$\log ((x - 1)y^2) = \log 9$

Similarly, the second equation becomes:

$\log (xy) = \log 5$

Since the logarithms are equal, we can equate the arguments (the expressions inside the logarithms). This gives us the following system of equations:

$\begin{cases} (x - 1)y^2 = 9 \\ xy = 5 \end{cases}$

From the second equation, we can express $x$ in terms of $y$ (or vice versa). Let's solve for $x$:

$x = \frac{5}{y}$

Now, substitute this expression for $x$ into the first equation:

$\left(\frac{5}{y} - 1\right)y^2 = 9$

Multiply through by $y^2$:

$(5 - y)y = 9$

Expanding and rearranging, we get a quadratic equation in $y$:

$5y - y^2 = 9$

$y^2 - 5y + 9 = 0$

To solve this quadratic equation, we can use the quadratic formula, which is given by:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -5$, and $c = 9$. Substituting these values, we get:

$y = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(9)}}{2(1)}$

$y = \frac{5 \pm \sqrt{25 - 36}}{2}$

$y = \frac{5 \pm \sqrt{-11}}{2}$

Since the discriminant (the value inside the square root) is negative, the solutions for $y$ are complex numbers. This indicates that there are no real solutions for $y$. However, let's proceed to analyze what this means for our original system of equations. The presence of complex solutions for $y$ implies that there are no real values of $y$ that satisfy the given system of logarithmic equations. Therefore, there are no real solutions for $x$ either, as $x$ is expressed in terms of $y$. In this context, understanding the nature of solutions and the domain of logarithmic functions is crucial. Logarithmic functions are defined only for positive arguments, and complex solutions fall outside the realm of real-number solutions typically sought in such problems. This exploration showcases the importance of not only applying formulas but also interpreting the results in the context of the original problem.

Conclusion

In this article, we explored two distinct yet interconnected mathematical concepts: finding the equation of a line perpendicular to a given line and solving systems of logarithmic equations. By understanding the relationships between slopes of perpendicular lines and utilizing the point-slope form, we can effectively determine the equation of a perpendicular line. Furthermore, mastering logarithmic properties and algebraic techniques allows us to solve systems of logarithmic equations, although, as demonstrated, not all systems have real solutions. These skills are fundamental in mathematics and have broad applications in various fields. Continuous practice and a deep understanding of the underlying principles are key to mastering these concepts and excelling in mathematical problem-solving. Through such exercises, we not only strengthen our mathematical abilities but also enhance our logical reasoning and analytical skills, which are invaluable assets in any field of study or profession.