Solving Logarithmic Equations A Step-by-Step Guide To Log(1/100) = Log(10^(x+2))

Introduction: Delving into the Realm of Logarithms

In the captivating world of mathematics, logarithms reign supreme as a powerful tool for simplifying complex equations and unraveling exponential relationships. Often perceived as enigmatic symbols, logarithms, in essence, are the inverse operations of exponentiation. This means that they elegantly answer the question: To what power must we raise a specific base to obtain a given number? In this comprehensive guide, we will embark on a journey to demystify logarithmic equations by meticulously dissecting and solving the equation log(1/100) = log(10^(x+2)). Along the way, we'll fortify your understanding of logarithmic properties and equip you with the problem-solving prowess to confidently conquer similar challenges.

Before we plunge into the depths of our equation, let's take a moment to grasp the fundamental concepts that underpin logarithms. A logarithm, at its core, is the exponent to which a base must be raised to produce a specific number. Symbolically, this is expressed as log_b(a) = c, where 'b' represents the base, 'a' is the number, and 'c' signifies the exponent. For instance, in the expression log_10(100) = 2, the base is 10, the number is 100, and the exponent is 2, which translates to 10 raised to the power of 2 equals 100. Grasping this fundamental relationship is the bedrock for navigating the world of logarithmic equations.

Deciphering the Equation: Unveiling the Logarithmic Puzzle

Now, let's cast our attention to the equation at hand: log(1/100) = log(10^(x+2)). This equation presents us with a fascinating challenge, an opportunity to wield our knowledge of logarithms to solve for the elusive variable 'x'. Our quest begins by recognizing that both sides of the equation boast logarithms. This pivotal observation opens the gateway to simplifying the equation by leveraging the inherent properties of logarithms. Specifically, we'll harness the power of the property that asserts if log_b(a) = log_b(c), then a = c. This property essentially states that if the logarithms of two numbers are equal and they share the same base, then the numbers themselves must be equal.

In our equation, the base of the logarithms is conspicuously absent, implying a base of 10, commonly known as the common logarithm. This revelation allows us to apply the aforementioned property, setting the stage for simplification. We can confidently equate the arguments of the logarithms, which are the expressions inside the logarithmic functions. This crucial step transforms our equation from a logarithmic realm into a more manageable algebraic landscape, paving the path for isolating and solving for 'x'. The arguments we're dealing with are 1/100 on the left side and 10^(x+2) on the right side. By equating these arguments, we effectively eliminate the logarithms, bringing us closer to the solution.

The Art of Simplification: Taming the Equation

With the logarithms vanquished, our equation now stands as 1/100 = 10^(x+2). This equation, while devoid of logarithms, still holds a subtle complexity. To unravel it, we need to express both sides of the equation using the same base. A keen observer will notice that 1/100 can be artfully rewritten as 10^(-2). This transformation is pivotal because it unveils the common base of 10, aligning both sides of the equation for a direct comparison of exponents. Remember, expressing numbers as powers of a common base is a cornerstone technique in simplifying exponential and logarithmic equations. It allows us to directly compare the exponents, paving the way for a streamlined solution.

Now, our equation elegantly transforms into 10^(-2) = 10^(x+2). With both sides proudly displaying the base of 10, we can invoke another fundamental property of exponential equations: if b^m = b^n, then m = n. This property empowers us to equate the exponents, thereby dissolving the exponential expressions and leaving us with a simple algebraic equation. In our case, this means equating -2 and (x+2), resulting in the equation -2 = x + 2. We've successfully navigated through the complexities of logarithms and exponents, arriving at a linear equation that's ripe for a straightforward solution. The journey of simplification has brought us closer to unveiling the value of 'x'.

Solving for the Unknown: Unveiling the Value of 'x'

Our equation, -2 = x + 2, now stands as a testament to our simplification efforts. It's a linear equation, a familiar terrain for algebraic manipulation. To isolate 'x', we employ the fundamental principle of maintaining equality: performing the same operation on both sides of the equation. In this case, we subtract 2 from both sides. This strategic move effectively cancels out the +2 on the right side, leaving 'x' isolated and vulnerable to discovery.

Subtracting 2 from both sides of the equation gracefully transforms it into -2 - 2 = x + 2 - 2, which further simplifies to -4 = x. Behold! The value of 'x' is unveiled: x = -4. Our algebraic expedition has reached its destination, revealing the solution to our equation. But our journey doesn't end here. It's crucial to verify our solution, ensuring its validity within the original equation. This act of verification is a hallmark of meticulous problem-solving, guarding against potential errors and solidifying our confidence in the result.

Verification: Ensuring the Solution's Integrity

To verify the solution x = -4, we embark on a critical step: substituting this value back into the original equation, log(1/100) = log(10^(x+2)). This substitution allows us to assess whether our solution holds true, whether it satisfies the equation's delicate balance. We replace 'x' with -4, transforming the right side of the equation into log(10^(-4+2)), which simplifies to log(10^(-2)).

Now, our equation reads log(1/100) = log(10^(-2)). To further validate our solution, we recall that 1/100 can be expressed as 10^(-2). This transformation allows us to rewrite the left side of the equation as log(10^(-2)). Lo and behold, both sides of the equation now stand in perfect harmony: log(10^(-2)) = log(10^(-2)). This resounding equality serves as irrefutable proof that our solution, x = -4, is indeed correct. The verification process not only confirms the accuracy of our solution but also deepens our understanding of the underlying logarithmic principles.

Conclusion: Mastering Logarithmic Equations

In this comprehensive exploration, we've successfully navigated the intricacies of the logarithmic equation log(1/100) = log(10^(x+2)), unveiling the solution x = -4. Our journey has been a testament to the power of logarithmic properties, algebraic manipulation, and the crucial role of verification. By understanding the fundamental concepts of logarithms, applying simplification techniques, and diligently verifying our solutions, we've honed our problem-solving skills and solidified our mastery of logarithmic equations.

Logarithms are not mere mathematical symbols; they are keys that unlock the doors to solving complex problems across diverse fields, from science and engineering to finance and computer science. As you continue your mathematical journey, embrace the power of logarithms, and let them guide you towards unraveling the mysteries of the numerical world. Remember, practice is the cornerstone of mastery. Challenge yourself with a variety of logarithmic equations, and watch your problem-solving prowess flourish. The world of logarithms awaits your exploration, promising a wealth of mathematical discoveries.

Therefore, the correct answer is C) x = -4.