Solving Logarithmic Equations A Step-by-Step Guide

Introduction

In the realm of mathematics, solving logarithmic equations is a fundamental skill that finds applications in various fields, including physics, engineering, and computer science. This article delves into the step-by-step process of solving the logarithmic equation ${\log_{10}(x-15) = 2 - \log_{10}x}$. We will explore the underlying principles, properties of logarithms, and algebraic manipulations required to arrive at the solution. Understanding these concepts is crucial for anyone looking to enhance their mathematical proficiency and tackle more complex problems. This guide aims to provide a clear and concise explanation, ensuring that readers can confidently solve similar equations in the future. By breaking down each step, we will illustrate how to effectively use logarithmic identities and algebraic techniques to find the solution, making the process accessible and understandable for all.

Understanding Logarithmic Equations

Before diving into the solution, it’s essential to grasp the basics of logarithmic equations. A logarithmic equation is an equation that involves a logarithm of an expression containing a variable. The logarithm, denoted as ${\log_b a}$, represents the power to which the base b must be raised to produce the number a. In the given equation, ${\log_{10}(x-15) = 2 - \log_{10}x}$, we are dealing with base-10 logarithms, also known as common logarithms. The key property we’ll use here is the logarithmic identity that allows us to combine logarithms using addition and subtraction. Specifically, ${\log_b m + \log_b n = \log_b(mn)}$ and ${\log_b m - \log_b n = \log_b(\frac{m}{n})}$. Understanding these properties is crucial as they allow us to manipulate and simplify logarithmic equations, making them easier to solve. Moreover, the definition of a logarithm helps us to convert logarithmic forms into exponential forms, which can be a pivotal step in solving such equations. Logarithmic equations are prevalent in various scientific and engineering contexts, making it important to understand their behavior and methods for solving them. Familiarity with the properties and techniques discussed will not only help in solving this specific equation but also in tackling a wide range of logarithmic problems.

Step-by-Step Solution

Let's solve the equation ${\log_{10}(x-15) = 2 - \log_{10}x}$ step by step. First, we need to combine the logarithmic terms on one side of the equation. To do this, we add ${\log_{10}x}$ to both sides of the equation:

${ \log_{10}(x-15) + \log_{10}x = 2 }$

Next, we use the logarithmic property ${\log_b m + \log_b n = \log_b(mn)}$ to combine the logarithms:

${ \log_{10}((x-15)x) = 2 }$

This simplifies to:

${ \log_{10}(x^2 - 15x) = 2 }$

Now, we convert the logarithmic equation into its exponential form. Since we are using base-10 logarithms, we rewrite the equation as:

${ 10^2 = x^2 - 15x }$

Which simplifies to:

${ 100 = x^2 - 15x }$

Rearranging the equation into a quadratic form gives us:

${ x^2 - 15x - 100 = 0 }$

This quadratic equation can be solved by factoring, completing the square, or using the quadratic formula. We will proceed with factoring.

Solving the Quadratic Equation

Now that we have the quadratic equation ${x^2 - 15x - 100 = 0}$, we need to solve it to find the possible values of x. We look for two numbers that multiply to -100 and add up to -15. These numbers are -20 and 5. Thus, we can factor the quadratic equation as follows:

${ (x - 20)(x + 5) = 0 }$

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

${ x - 20 = 0 \Rightarrow x = 20 }$

${ x + 5 = 0 \Rightarrow x = -5 }$

So, our potential solutions are ${x = 20}$ and ${x = -5}$. However, we must check these solutions in the original logarithmic equation to ensure they are valid, as logarithms are only defined for positive arguments.

Checking for Extraneous Solutions

When solving logarithmic equations, it's crucial to check for extraneous solutions. These are solutions that satisfy the transformed equation but not the original equation. Logarithmic functions are only defined for positive arguments, so we must ensure that the values inside the logarithms are positive. Let's check our potential solutions, ${x = 20}$ and ${x = -5}$, in the original equation:

${ \log_{10}(x-15) = 2 - \log_{10}x }$

For ${x = 20}$:

${ \log_{10}(20-15) = 2 - \log_{10}20 }$

${ \log_{10}(5) = 2 - \log_{10}20 }$

Using the property ${\log_{10}20 = \log_{10}(10 \times 2) = \log_{10}10 + \log_{10}2 = 1 + \log_{10}2}$, we have:

${ \log_{10}(5) = 2 - (1 + \log_{10}2) }$

${ \log_{10}(5) = 1 - \log_{10}2 }$

${ \log_{10}(5) + \log_{10}2 = 1 }$

${ \log_{10}(5 \times 2) = 1 }$

${ \log_{10}(10) = 1 }$

This is true, so ${x = 20}$ is a valid solution.

Now, let's check ${x = -5}$:

${ \log_{10}(-5-15) = 2 - \log_{10}(-5) }$

Since we cannot take the logarithm of a negative number, ${x = -5}$ is an extraneous solution. Therefore, the only valid solution to the equation is ${x = 20}$.

Conclusion

In conclusion, the solution to the logarithmic equation ${\log_{10}(x-15) = 2 - \log_{10}x}$ is ${x = 20}$. We arrived at this solution by first combining the logarithmic terms, converting the equation to exponential form, and solving the resulting quadratic equation. We then checked for extraneous solutions to ensure the validity of our answer. This process highlights the importance of understanding the properties of logarithms and the need for careful verification in solving logarithmic equations. By following these steps, one can effectively solve similar logarithmic problems and enhance their mathematical toolkit. The skills acquired in solving such equations are invaluable in various scientific and engineering applications, making the effort to master these techniques well worth it. Remember to always check for extraneous solutions, as this step is crucial in ensuring the correctness of your results and avoiding common pitfalls in logarithmic equation solving.