Solving Log₅(y²+y) = Log₅(3y-1) A Step-by-Step Guide

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When confronted with logarithmic equations, the task of isolating the variable might seem daunting at first. However, with a systematic approach and a solid understanding of logarithmic properties, solving these equations can become a manageable endeavor. In this article, we will delve into the process of solving the equation log₅(y²+y) = log₅(3y-1), providing a detailed step-by-step solution while highlighting key concepts and potential pitfalls along the way. This exploration will not only equip you with the necessary skills to tackle similar problems but also enhance your overall understanding of logarithmic functions and their applications.

Understanding Logarithmic Equations

Before diving into the specifics of our equation, let's establish a firm grasp on the fundamental principles of logarithmic equations. A logarithmic equation is essentially an equation where the logarithm of an expression containing a variable is set equal to a constant or another logarithmic expression. The key to solving these equations lies in leveraging the properties of logarithms to simplify the equation and ultimately isolate the variable. The most crucial property we'll utilize in this case is the one-to-one property of logarithms. This property states that if logₐ(b) = logₐ(c), then b = c, provided that a, b, and c are positive and a ≠ 1. This property allows us to eliminate the logarithmic functions from the equation, transforming it into a more manageable algebraic form.

It's also important to remember the definition of a logarithm. The expression logₐ(b) = x is equivalent to aˣ = b. This definition provides the foundational link between logarithms and exponents, enabling us to switch between these forms as needed to solve equations. Furthermore, we must always be mindful of the domain of logarithmic functions. The argument of a logarithm (the expression inside the logarithm) must be strictly positive. This constraint is crucial when checking for extraneous solutions, which are solutions obtained algebraically but do not satisfy the original equation due to domain restrictions.

Step-by-Step Solution of log₅(y²+y) = log₅(3y-1)

Now, let's embark on the journey of solving the equation log₅(y²+y) = log₅(3y-1). We will meticulously dissect each step, providing clear explanations and justifications to ensure a thorough understanding of the process.

Step 1: Apply the One-to-One Property of Logarithms

The cornerstone of our solution lies in the one-to-one property of logarithms. Since we have logarithms with the same base (base 5) on both sides of the equation, we can confidently equate the arguments of the logarithms. This gives us:

y² + y = 3y - 1

This step effectively eliminates the logarithmic functions, transforming the equation into a quadratic equation, which we are well-equipped to handle.

Step 2: Rearrange the Equation into Standard Quadratic Form

To solve the quadratic equation, we need to rearrange it into the standard form, which is ax² + bx + c = 0. Subtracting 3y and adding 1 to both sides of the equation, we get:

y² + y - 3y + 1 = 0

Simplifying, we obtain:

y² - 2y + 1 = 0

Now, the equation is in the familiar standard quadratic form, ready for the next step in our solution.

Step 3: Solve the Quadratic Equation

We have several options for solving the quadratic equation y² - 2y + 1 = 0. We can attempt to factor it, use the quadratic formula, or complete the square. In this case, the equation is readily factorable. We recognize that the left-hand side is a perfect square trinomial:

(y - 1)² = 0

Taking the square root of both sides, we get:

y - 1 = 0

Adding 1 to both sides, we find:

y = 1

Thus, we have obtained a potential solution for our equation. However, we must always remember the crucial step of checking for extraneous solutions.

Step 4: Check for Extraneous Solutions

The domain of a logarithmic function is restricted to positive arguments. Therefore, we must ensure that the value of y we found, y = 1, does not result in negative or zero arguments within the original logarithmic equation. Let's substitute y = 1 into the original equation:

log₅(y² + y) = log₅(3y - 1)

log₅(1² + 1) = log₅(3(1) - 1)

log₅(2) = log₅(2)

Since both arguments, y² + y = 2 and 3y - 1 = 2, are positive when y = 1, our solution is valid. There are no extraneous solutions in this case.

Step 5: State the Solution

Having successfully solved the equation and verified the solution, we can confidently state that the solution to the equation log₅(y² + y) = log₅(3y - 1) is:

y = 1

Common Pitfalls and How to Avoid Them

Solving logarithmic equations can be tricky, and there are several common pitfalls that students often encounter. Being aware of these pitfalls and understanding how to avoid them is crucial for achieving accurate solutions.

Forgetting to Check for Extraneous Solutions

This is perhaps the most common mistake. As emphasized earlier, the arguments of logarithmic functions must be positive. Failing to check for extraneous solutions can lead to incorrect answers. Always substitute your solutions back into the original equation to ensure that they do not violate the domain restrictions.

Incorrectly Applying Logarithmic Properties

Logarithmic properties, such as the product rule, quotient rule, and power rule, are powerful tools for simplifying logarithmic expressions and equations. However, they must be applied correctly. For instance, logₐ(b + c) is not equal to logₐ(b) + logₐ(c). A thorough understanding of these properties is essential for accurate manipulation of logarithmic expressions.

Making Algebraic Errors

Solving logarithmic equations often involves algebraic manipulations, such as rearranging terms, factoring, and solving quadratic equations. Careless algebraic errors can lead to incorrect solutions. Double-check your work, paying close attention to signs and operations.

Misunderstanding the Domain of Logarithmic Functions

The domain of a logarithmic function is restricted to positive arguments. This means that the expression inside the logarithm must be greater than zero. Misunderstanding this restriction can lead to incorrect solutions and a failure to identify extraneous solutions.

Real-World Applications of Logarithmic Equations

Logarithmic equations are not merely abstract mathematical concepts; they have a wide range of real-world applications across various fields. Understanding these applications can provide a deeper appreciation for the importance of logarithmic equations.

Exponential Growth and Decay

Logarithmic equations are fundamental in modeling phenomena involving exponential growth and decay. For example, they are used to determine the time it takes for an investment to double at a given interest rate, or to calculate the half-life of a radioactive substance. These applications rely on the inverse relationship between exponential and logarithmic functions.

Richter Scale

The Richter scale, used to measure the magnitude of earthquakes, is a logarithmic scale. Each whole number increase on the Richter scale represents a tenfold increase in the amplitude of the seismic waves. Logarithmic equations are used to convert seismic measurements into Richter scale values.

Sound Intensity

The decibel scale, used to measure sound intensity, is also a logarithmic scale. The loudness of a sound is perceived logarithmically by the human ear. Logarithmic equations are used to relate sound intensity to decibel levels.

Chemical Reactions

In chemistry, logarithmic equations are used to describe the rate of chemical reactions and to determine the pH of solutions. The pH scale is a logarithmic scale that measures the acidity or alkalinity of a solution.

Conclusion

Solving logarithmic equations requires a systematic approach, a solid understanding of logarithmic properties, and meticulous attention to detail. By applying the one-to-one property of logarithms, rearranging equations into standard forms, and carefully checking for extraneous solutions, we can effectively tackle these equations. The equation log₅(y² + y) = log₅(3y - 1) serves as a prime example of this process, showcasing the importance of each step. Furthermore, recognizing common pitfalls and understanding the real-world applications of logarithmic equations enhances our overall comprehension and appreciation of this mathematical concept. Mastering the art of solving logarithmic equations not only strengthens our mathematical prowess but also equips us with valuable tools for tackling problems in various scientific and engineering disciplines.

By diligently practicing and applying these techniques, you can confidently navigate the world of logarithmic equations and unlock their potential in solving a wide array of problems.