Solving Exponential Equations 2^(y^2-4y)=32 Step-by-Step Guide
Exponential equations can seem daunting, but with a systematic approach, you can conquer even the most complex ones. In this comprehensive guide, we'll break down the process of solving the exponential equation 2(y2-4y) = 32, providing you with a clear understanding of the steps involved and the underlying principles. Whether you're a student tackling homework or a math enthusiast looking to expand your skills, this guide will equip you with the knowledge and confidence to solve exponential equations effectively.
Understanding Exponential Equations
Before we dive into the specifics of solving 2(y2-4y) = 32, let's establish a solid foundation by understanding what exponential equations are and their key properties. An exponential equation is an equation in which the variable appears in the exponent. These equations often model real-world phenomena like population growth, radioactive decay, and compound interest, making them crucial in various scientific and financial applications.
The general form of an exponential equation is a^(f(x)) = b, where 'a' is the base, 'f(x)' is the exponent (a function of the variable 'x'), and 'b' is a constant. The key to solving exponential equations lies in manipulating the equation to isolate the variable. This often involves using the properties of exponents and logarithms.
One of the fundamental properties we'll use is that if a^m = a^n, then m = n. This property allows us to equate the exponents when the bases are the same, simplifying the equation significantly. Another crucial concept is the relationship between exponential and logarithmic functions. The logarithm is the inverse operation of exponentiation. Therefore, if a^x = y, then log_a(y) = x. This relationship is particularly useful when we cannot easily express both sides of the equation with the same base.
In the case of our equation, 2(y2-4y) = 32, we can see that the base is 2. Our goal is to express 32 as a power of 2, which will allow us to equate the exponents and solve for 'y'. This initial step of recognizing and manipulating the bases is often the most critical part of solving exponential equations.
Step 1: Express Both Sides with the Same Base
The first crucial step in solving the equation 2(y2-4y) = 32 is to express both sides of the equation with the same base. This allows us to leverage the property that if a^m = a^n, then m = n. In our case, the left side of the equation already has a base of 2. Therefore, we need to express 32 as a power of 2.
We can do this by recognizing that 32 is a power of 2. Specifically, 32 is equal to 2 multiplied by itself five times: 32 = 2 * 2 * 2 * 2 * 2 = 2^5. This is a fundamental step and often involves knowing the powers of common numbers like 2, 3, and 5. Recognizing these powers can significantly speed up the solving process.
Now, we can rewrite our original equation 2(y2-4y) = 32 as 2(y2-4y) = 2^5. By expressing both sides of the equation with the same base, we have set the stage for the next step, which involves equating the exponents.
This step is crucial because it transforms the exponential equation into a more manageable algebraic equation. By ensuring that both sides of the equation have the same base, we can eliminate the exponential part and focus on solving the resulting polynomial equation. Without this step, solving the equation would be considerably more challenging.
Understanding and applying this step correctly is essential for mastering the solution of exponential equations. It not only simplifies the equation at hand but also lays a foundation for solving more complex problems in the future. The ability to recognize and manipulate powers of numbers is a fundamental skill in algebra and is invaluable in various mathematical contexts.
Step 2: Equate the Exponents
Having successfully expressed both sides of the equation 2(y2-4y) = 32 with the same base, we now have 2(y2-4y) = 2^5. The next crucial step is to equate the exponents. This is based on the fundamental property of exponential functions: if a^m = a^n, then m = n. This property is the cornerstone of solving exponential equations where the bases are the same.
Applying this property to our equation, we can equate the exponents, which gives us the quadratic equation: y^2 - 4y = 5. This step is significant because it transforms the exponential equation into a quadratic equation, which is a familiar type of algebraic equation that we can solve using various methods, such as factoring, completing the square, or using the quadratic formula.
The process of equating exponents simplifies the problem considerably. Instead of dealing with an exponential function, we now have a polynomial equation to solve. This simplification is a key strategy in solving exponential equations and demonstrates the power of using the properties of exponents to our advantage.
This step not only makes the equation solvable but also provides a clear path forward. By reducing the exponential equation to a quadratic equation, we can apply our knowledge of algebra to find the values of 'y' that satisfy the equation. The transition from exponential to polynomial form is a critical skill in mathematics and is used in many different contexts.
Therefore, the step of equating the exponents is a pivotal moment in solving exponential equations. It transforms the problem into a more accessible form, allowing us to use standard algebraic techniques to find the solution. This step highlights the importance of understanding and applying the fundamental properties of exponents.
Step 3: Solve the Quadratic Equation
After equating the exponents in the equation 2(y2-4y) = 32, we arrived at the quadratic equation y^2 - 4y = 5. Now, the task is to solve this quadratic equation for 'y'. Quadratic equations are of the form ax^2 + bx + c = 0, and there are several methods to solve them, including factoring, completing the square, and using the quadratic formula. In this case, factoring is the most straightforward approach.
First, we need to rewrite the equation in the standard quadratic form by subtracting 5 from both sides: y^2 - 4y - 5 = 0. This step is crucial because it sets the equation equal to zero, which is necessary for factoring.
Next, we look for two numbers that multiply to -5 and add to -4. These numbers are -5 and 1. Therefore, we can factor the quadratic equation as follows: (y - 5)(y + 1) = 0. Factoring is a powerful technique for solving quadratic equations, as it breaks down the equation into simpler factors.
Once we have the factored form, we can apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This gives us two separate equations: y - 5 = 0 and y + 1 = 0.
Solving each of these equations for 'y', we find two potential solutions: y = 5 and y = -1. These are the values of 'y' that make the quadratic equation true. However, it's essential to verify these solutions in the original exponential equation to ensure they are valid.
Solving the quadratic equation is a critical step in finding the solutions to the original exponential equation. The ability to factor quadratic equations or use other methods like the quadratic formula is a fundamental skill in algebra and is applied in many different areas of mathematics. By solving the quadratic equation, we have identified the potential values of 'y' that satisfy the exponential equation.
Step 4: Verify the Solutions
Having found the potential solutions y = 5 and y = -1 from solving the quadratic equation y^2 - 4y - 5 = 0, the final crucial step is to verify these solutions in the original exponential equation 2(y2-4y) = 32. This step is essential because sometimes solutions obtained from solving transformed equations may not satisfy the original equation due to extraneous roots or other algebraic manipulations.
Let's start by verifying y = 5. Substitute y = 5 into the original equation: 2((5)2-4(5)) = 2^(25-20) = 2^5 = 32. Since this equals the right side of the original equation, y = 5 is a valid solution.
Next, we verify y = -1. Substitute y = -1 into the original equation: 2((-1)2-4(-1)) = 2^(1+4) = 2^5 = 32. This also equals the right side of the original equation, so y = -1 is also a valid solution.
Both potential solutions satisfy the original exponential equation. Therefore, we can confidently conclude that the solutions to the equation 2(y2-4y) = 32 are y = 5 and y = -1.
Verifying solutions is a critical practice in solving any equation, whether it's exponential, quadratic, or any other type. It ensures that the solutions we find are indeed correct and that we haven't made any errors in our algebraic manipulations. This step provides a final check and adds confidence to our answer.
In this case, both solutions were valid, but in other problems, you might encounter extraneous solutions that do not satisfy the original equation. Therefore, always remember to verify your solutions to ensure accuracy and completeness.
Final Answer
After systematically solving the exponential equation 2(y2-4y) = 32, we have arrived at the final answer. By expressing both sides of the equation with the same base, equating the exponents, solving the resulting quadratic equation, and verifying the solutions, we have determined that the values of 'y' that satisfy the equation are y = 5 and y = -1.
Therefore, the final answer is y = 5, -1.
This comprehensive solution demonstrates the step-by-step process of solving exponential equations. It highlights the importance of understanding the properties of exponents, the techniques for solving quadratic equations, and the necessity of verifying solutions. By following this systematic approach, you can confidently tackle similar exponential equations.
This example showcases the power of mathematical principles in solving real problems. The combination of exponential and quadratic equation solving techniques is a testament to the interconnectedness of mathematical concepts. Mastering these techniques will not only help you in academic settings but also in various practical applications where exponential models are used.
In conclusion, the solutions to the exponential equation 2(y2-4y) = 32 are y = 5 and y = -1. By following a methodical approach and understanding the underlying mathematical principles, you can successfully solve a wide range of exponential equations.
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