Solving Exponential Equations For X A Comprehensive Guide

by Scholario Team 58 views

In the realm of mathematics, exponential equations hold a significant position, often appearing in various contexts, from scientific calculations to financial modeling. Understanding how to solve these equations is crucial for anyone seeking to master mathematical concepts. This article delves into a specific exponential equation, providing a step-by-step guide to finding the value of the unknown variable, x. We will explore the fundamental principles governing exponential equations and apply them to solve the equation:

(75)x×(75)−5=(75)2x−10\left(\frac{7}{5}\right)^x \times \left(\frac{7}{5}\right)^{-5}=\left(\frac{7}{5}\right)^{2 x-10}

Join us as we unravel the intricacies of this equation and equip you with the knowledge to tackle similar problems with confidence. We'll break down the equation, explain the reasoning behind each step, and provide insights into the broader context of exponential equations. By the end of this guide, you'll not only know the solution to this particular equation but also have a solid understanding of the techniques involved in solving exponential equations in general.

Before diving into the solution, it's essential to grasp the fundamental principles of exponential equations. These equations involve variables in the exponents, making them distinct from algebraic equations where variables are in the base. The key to solving exponential equations lies in manipulating them to isolate the variable. One of the core concepts is the rule of exponents, which states that when multiplying exponential terms with the same base, you can add the exponents. This rule is the cornerstone of our approach to solving the given equation.

The equation we're tackling is a classic example of an exponential equation where the base is a fraction, and the exponents involve the variable x. To successfully solve this, we'll need to apply the rules of exponents judiciously and simplify the equation step by step. By understanding the underlying principles, you'll be able to approach similar problems with a clear strategy and avoid common pitfalls. The goal is to transform the equation into a form where we can directly compare the exponents and solve for x. This involves combining terms, simplifying expressions, and ultimately isolating the variable.

Now, let's embark on the journey of solving the equation. We'll break down each step and provide explanations to ensure clarity. Our primary goal is to isolate x, and we'll achieve this by applying the rules of exponents and simplifying the equation progressively.

Step 1: Applying the Product of Powers Rule

The first step involves applying the product of powers rule, which states that when multiplying exponential terms with the same base, we add the exponents. In our equation,

(75)x×(75)−5=(75)2x−10\left(\frac{7}{5}\right)^x \times \left(\frac{7}{5}\right)^{-5}=\left(\frac{7}{5}\right)^{2 x-10}

we have two terms on the left side with the same base (7/5). Applying the product of powers rule, we add the exponents x and -5:

(75)x+(−5)=(75)2x−10\left(\frac{7}{5}\right)^{x + (-5)} = \left(\frac{7}{5}\right)^{2 x-10}

This simplifies to:

(75)x−5=(75)2x−10\left(\frac{7}{5}\right)^{x - 5} = \left(\frac{7}{5}\right)^{2 x-10}

This step is crucial because it combines the two terms on the left side into a single exponential term, making the equation easier to handle. By adding the exponents, we've reduced the complexity and set the stage for the next step in the solution.

Step 2: Equating the Exponents

Now that we have a single exponential term on each side of the equation with the same base, we can equate the exponents. This is a fundamental property of exponential equations: if a^m = a^n, then m = n. Applying this principle to our equation,

(75)x−5=(75)2x−10\left(\frac{7}{5}\right)^{x - 5} = \left(\frac{7}{5}\right)^{2 x-10}

we can equate the exponents:

x−5=2x−10x - 5 = 2x - 10

This step is a turning point in the solution process. By equating the exponents, we've transformed the exponential equation into a linear equation, which is much easier to solve. This step relies on the fact that exponential functions are one-to-one, meaning that each input corresponds to a unique output, and vice versa. Therefore, if the bases are the same, the exponents must be equal for the equation to hold true.

Step 3: Solving for x

We now have a linear equation, which we can solve using standard algebraic techniques. Our equation is:

x−5=2x−10x - 5 = 2x - 10

To solve for x, we need to isolate it on one side of the equation. Let's start by subtracting x from both sides:

x−5−x=2x−10−xx - 5 - x = 2x - 10 - x

This simplifies to:

−5=x−10-5 = x - 10

Next, we add 10 to both sides:

−5+10=x−10+10-5 + 10 = x - 10 + 10

This simplifies to:

5=x5 = x

Therefore, the value of x is 5. This step involves basic algebraic manipulations to isolate the variable. By performing the same operations on both sides of the equation, we maintain the equality and gradually work towards isolating x. The result, x = 5, is the solution to the linear equation and, consequently, the solution to the original exponential equation.

Step 4: Verifying the Solution

To ensure our solution is correct, it's always a good practice to verify it by substituting the value of x back into the original equation. Our original equation was:

(75)x×(75)−5=(75)2x−10\left(\frac{7}{5}\right)^x \times \left(\frac{7}{5}\right)^{-5}=\left(\frac{7}{5}\right)^{2 x-10}

Substituting x = 5, we get:

(75)5×(75)−5=(75)2(5)−10\left(\frac{7}{5}\right)^5 \times \left(\frac{7}{5}\right)^{-5}=\left(\frac{7}{5}\right)^{2(5)-10}

Simplifying the right side:

(75)2(5)−10=(75)10−10=(75)0\left(\frac{7}{5}\right)^{2(5)-10} = \left(\frac{7}{5}\right)^{10-10} = \left(\frac{7}{5}\right)^0

Any number raised to the power of 0 is 1, so:

(75)0=1\left(\frac{7}{5}\right)^0 = 1

Now, let's simplify the left side using the product of powers rule:

(75)5×(75)−5=(75)5+(−5)=(75)0\left(\frac{7}{5}\right)^5 \times \left(\frac{7}{5}\right)^{-5} = \left(\frac{7}{5}\right)^{5 + (-5)} = \left(\frac{7}{5}\right)^0

Again, any number raised to the power of 0 is 1, so:

(75)0=1\left(\frac{7}{5}\right)^0 = 1

Since both sides of the equation equal 1 when x = 5, our solution is verified. This step is crucial for confirming the accuracy of the solution. By substituting the value back into the original equation, we ensure that the equality holds true. If the equation doesn't balance, it indicates an error in the solution process, and we need to revisit the steps to identify the mistake.

In this article, we've successfully solved the exponential equation

(75)x×(75)−5=(75)2x−10\left(\frac{7}{5}\right)^x \times \left(\frac{7}{5}\right)^{-5}=\left(\frac{7}{5}\right)^{2 x-10}

and found the value of x to be 5. We accomplished this by applying the product of powers rule, equating exponents, and solving the resulting linear equation. The verification step further solidified our confidence in the solution.

Understanding exponential equations is vital for various mathematical applications. The techniques we've discussed here can be applied to a wide range of similar problems. Remember, the key is to simplify the equation using the rules of exponents, transform it into a manageable form, and then solve for the variable. By mastering these skills, you'll be well-equipped to tackle more complex mathematical challenges.

This exploration has not only provided a solution to a specific equation but also illuminated the broader landscape of exponential equations. With a firm grasp of the principles and techniques discussed, you're now better prepared to navigate the world of mathematical problem-solving. Keep practicing, and you'll continue to enhance your skills and deepen your understanding.

Solve for x: (7/5)^x * (7/5)^(-5) = (7/5)^(2x-10)

Solving for x in Exponential Equations A Step-by-Step Guide