Solving E^(2x) - 10e^x + 24 = 0: Find Both Solutions

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Solving the Exponential Equation e^(2x) - 10e^x + 24 = 0

Hey everyone! Let's dive into solving this interesting exponential equation: e^(2x) - 10e^x + 24 = 0. It looks a bit intimidating at first, but don't worry, we'll break it down step-by-step. Our goal is to find both the smaller and larger solutions for x. So, grab your thinking caps, and let's get started!

Understanding the Equation

First, let's understand what we're dealing with. This equation involves exponential terms, specifically e raised to the power of x and 2x. The key here is to recognize that e^(2x) can be rewritten as (ex)2. This simple transformation is crucial because it allows us to turn the equation into a more familiar form. By recognizing the structure of the equation, we can apply algebraic techniques that are typically used for solving quadratic equations. This involves making a substitution that simplifies the equation, making it easier to manipulate and solve. The ability to transform equations into more manageable forms is a fundamental skill in mathematics, and it is particularly useful in dealing with exponential and logarithmic functions.

When we spot this, we can make a substitution to simplify things. Let's set y = e^x. Now, our equation transforms into a quadratic equation in terms of y. This is a classic technique in algebra – turning a complex equation into a simpler one that we know how to solve. This method of substitution not only simplifies the algebraic manipulation but also provides a clearer pathway to the solution. It's like changing the lens through which we view the problem, revealing an underlying structure that is easier to handle. Remember, mathematics is often about finding the right perspective to make a problem tractable. Recognizing these patterns and applying appropriate substitutions is a cornerstone of mathematical problem-solving.

Transforming into a Quadratic Equation

With our substitution y = e^x, the equation e^(2x) - 10e^x + 24 = 0 becomes y^2 - 10y + 24 = 0. See? Much simpler! Now we have a standard quadratic equation that we can solve using several methods, such as factoring, completing the square, or the quadratic formula. Factoring is often the quickest method if you can spot the factors easily, while the quadratic formula is a more general approach that works for any quadratic equation. Completing the square is another powerful technique that not only solves quadratic equations but also provides insights into the structure of the equation and its solutions. Each method has its advantages, and choosing the right one often depends on the specific equation and your personal preference.

Solving the Quadratic Equation

Let's solve y^2 - 10y + 24 = 0 by factoring. We need to find two numbers that multiply to 24 and add up to -10. Those numbers are -6 and -4. So, we can factor the equation as (y - 6)(y - 4) = 0. This means that either y - 6 = 0 or y - 4 = 0. Therefore, the solutions for y are y = 6 and y = 4. Factoring is a powerful technique that relies on recognizing patterns in the equation. It's a bit like reverse-engineering the multiplication process, and it often leads to a quick solution. The ability to factor quadratic expressions efficiently is a valuable skill in algebra, and it comes with practice and familiarity with different types of factoring patterns. Remember, factoring is not just about finding the right numbers; it's about understanding the underlying structure of the equation and how it can be decomposed into simpler parts.

Back to the Original Variable

Okay, we've found the values for y, but remember, we want the values for x. We know that y = e^x, so we need to substitute back. This gives us two equations:

  1. e^x = 6
  2. e^x = 4

To solve for x, we need to use the natural logarithm (ln), which is the inverse function of e^x. Applying the natural logarithm to both sides of each equation will isolate x. The natural logarithm is a fundamental function in calculus and mathematical analysis, and it is essential for working with exponential equations. It allows us to "undo" the exponentiation and bring the variable down to a level where we can solve for it directly. The natural logarithm has many important properties that make it a versatile tool in various mathematical contexts, including solving differential equations and modeling growth and decay phenomena.

Solving for x

Taking the natural logarithm of both sides of e^x = 6, we get x = ln(6). Similarly, for e^x = 4, we get x = ln(4). These are our two solutions for x. Now, we just need to figure out which one is smaller and which is larger. Since the natural logarithm is an increasing function, the larger the number, the larger its natural logarithm. Therefore, ln(6) is greater than ln(4). Understanding the properties of functions, such as monotonicity (whether a function is increasing or decreasing), is crucial for comparing values and making informed decisions about the solutions. In this case, the increasing nature of the natural logarithm allows us to directly compare the solutions based on the values inside the logarithm.

Identifying the Smaller and Larger Solutions

So, the smaller solution is x = ln(4), and the larger solution is x = ln(6). We've successfully solved the equation and found both solutions! Remember, the natural logarithm of a number between 0 and 1 is negative, the natural logarithm of 1 is 0, and the natural logarithm of a number greater than 1 is positive. This behavior is a direct consequence of the properties of the exponential function and its inverse relationship with the natural logarithm. Understanding these fundamental properties is essential for working with logarithms and exponential functions effectively.

Final Answer

The smaller solution is ln(4), and the larger solution is ln(6). Great job, guys! We tackled a tricky exponential equation by using substitution, factoring, and natural logarithms. Keep practicing, and you'll become a pro at solving these types of problems. Remember, mathematics is a journey of exploration and discovery, and every problem you solve is a step forward on that journey.

To recap, here's what we did to solve the equation e^(2x) - 10e^x + 24 = 0:

  1. Recognized the form: We identified that e^(2x) could be written as (ex)2.
  2. Substituted: We let y = e^x to transform the equation into a quadratic equation.
  3. Solved the quadratic: We factored the quadratic equation y^2 - 10y + 24 = 0 to find y = 4 and y = 6.
  4. Substituted back: We replaced y with e^x to get e^x = 4 and e^x = 6.
  5. Used natural logarithms: We took the natural logarithm of both sides of each equation to solve for x.
  6. Identified solutions: We found the solutions x = ln(4) and x = ln(6).
  7. Determined smaller and larger: We identified ln(4) as the smaller solution and ln(6) as the larger solution.
  • Practice makes perfect: The more you practice solving exponential equations, the more comfortable you'll become with the techniques involved. Try solving similar problems with different coefficients and constants to build your skills.
  • Master the fundamentals: Make sure you have a solid understanding of exponential functions, logarithms, and quadratic equations. These are the building blocks for solving more complex problems.
  • Look for patterns: Train yourself to recognize patterns in equations. This will help you choose the most efficient method for solving them.
  • Check your answers: Always check your solutions by plugging them back into the original equation to make sure they are correct. This will help you avoid mistakes and build confidence in your problem-solving abilities.
  • Use online resources: There are many excellent online resources available to help you learn and practice solving exponential equations. Websites like Khan Academy, Wolfram Alpha, and Symbolab offer tutorials, examples, and practice problems.

Solving exponential equations might seem challenging at first, but with a systematic approach and a good understanding of the underlying concepts, you can master them. Remember to break down the problem into smaller steps, use substitutions when appropriate, and apply the properties of logarithms and exponential functions. Keep practicing, and you'll be solving complex equations in no time! Happy problem-solving, guys!