Solving E^(2x) - 5e^x + 6 = 0: Natural Logarithm Solution
Hey guys! Let's dive into a fun math problem today. We're going to solve the equation e^(2x) - 5e^x + 6 = 0, and we're going to express our solution in terms of natural logarithms. If you're scratching your head already, don't worry! We'll break it down step by step so it's super easy to follow. So grab your thinking caps, and let’s get started!
Understanding the Equation
Before we jump into solving this exponential equation, let's make sure we understand what we're dealing with. The equation e^(2x) - 5e^x + 6 = 0 looks a bit intimidating at first glance, but it's actually quite manageable once we recognize its structure. The key here is to notice that we have a term with e raised to the power of 2x and another term with e raised to the power of x. This kind of structure often suggests a clever substitution to make things simpler. Think of it like this: if we let y = e^x, then e^(2x) becomes (e^x)^2, which is just y^2. This substitution will transform our exponential equation into a quadratic equation, which we already know how to solve!
Why is this helpful? Well, quadratic equations have a standard form that we can easily work with. They look like ay^2 + by + c = 0, where a, b, and c are constants. We have several methods to solve these, such as factoring, completing the square, or using the quadratic formula. By transforming our original equation into a quadratic equation, we unlock a whole toolbox of techniques that we can use to find the solution. This approach is a classic example of how mathematical problem-solving often involves recognizing patterns and using transformations to simplify complex problems. Once we solve for y, we'll just need to remember to substitute back e^x for y and then solve for x using natural logarithms. So, let's get to it and see how this substitution works in practice!
Substitution: The Magic Trick
So, as mentioned earlier, let’s use a little substitution trick to simplify our lives. We're going to let y = e^x. This might seem like a random move, but trust me, it’s going to make things much clearer. Now, if y = e^x, then y^2 is just (e^x)^2, which simplifies to e^(2x). See where we're going with this? By making this substitution, we can rewrite our original equation in terms of y instead of e^x.
Our equation e^(2x) - 5e^x + 6 = 0 now transforms into y^2 - 5y + 6 = 0. Isn't that much nicer to look at? We've turned a somewhat scary exponential equation into a friendly quadratic equation. This is a classic technique in mathematics: transforming a problem into a form that we already know how to solve. Now we've got a quadratic equation in the standard form ay^2 + by + c = 0, where a = 1, b = -5, and c = 6. This quadratic equation is something we can easily handle using several methods, such as factoring, the quadratic formula, or completing the square. The beauty of this substitution is that it allows us to apply familiar techniques to a new type of problem. So, let's move on to the next step and solve this quadratic equation for y. We're well on our way to finding the solution for x!
Solving the Quadratic Equation
Alright, now that we've transformed our exponential equation into a quadratic equation, y^2 - 5y + 6 = 0, it's time to solve for y. There are a few ways we can tackle this, but let's go with factoring because it’s often the quickest method when it works. Factoring involves finding two numbers that multiply to give the constant term (6 in this case) and add up to give the coefficient of the y term (-5 in this case).
Think about it: what two numbers multiply to 6 and add up to -5? Well, -2 and -3 fit the bill perfectly! So, we can factor the quadratic equation as follows: (y - 2)(y - 3) = 0. This is where the magic of factoring comes in. If the product of two factors is zero, then at least one of the factors must be zero. This gives us two possible solutions for y: either y - 2 = 0 or y - 3 = 0. Solving these simple equations, we find that y = 2 or y = 3. Great! We've found the values of y that satisfy our quadratic equation. But remember, we're not ultimately interested in y; we want to find the values of x that solve our original exponential equation. So, the next step is to substitute back e^x for y and solve for x. We're getting closer to the final answer!
Factoring Method
As we just discussed, we can solve the quadratic equation y^2 - 5y + 6 = 0 by factoring. We look for two numbers that multiply to 6 (the constant term) and add up to -5 (the coefficient of the y term). Those numbers are -2 and -3. Therefore, we can write the quadratic equation as (y - 2)(y - 3) = 0. This factored form is super helpful because it allows us to easily find the solutions for y. If the product of two factors is zero, then at least one of the factors must be zero.
So, we have two possibilities: either y - 2 = 0 or y - 3 = 0. If y - 2 = 0, then adding 2 to both sides gives us y = 2. If y - 3 = 0, then adding 3 to both sides gives us y = 3. These are our two solutions for y: y = 2 and y = 3. Factoring is a powerful technique because it breaks down a complex problem into simpler parts. In this case, it allowed us to transform a quadratic equation into two linear equations, which are much easier to solve. Now that we've found the values of y, we're ready to take the next step and substitute back to find the values of x. Remember, the goal is to solve for x in the original exponential equation, so we're not quite done yet!
Back to the Original Variable
Okay, awesome! We've found that y = 2 or y = 3. But hold on, we were originally trying to solve for x, not y. Remember our substitution? We said that y = e^x. So, now we need to substitute e^x back in for y to find the values of x. This means we have two equations to solve:
e^x = 2e^x = 3
These are much simpler equations to deal with than the original one, aren't they? Now, how do we solve for x when it's stuck up there in the exponent? This is where logarithms come to the rescue! Specifically, we're going to use the natural logarithm, which is the logarithm with base e. The natural logarithm is the inverse function of the exponential function with base e, so it's perfectly suited for this situation. The natural logarithm is written as ln(x). The key property we'll use is that if e^x = a, then ln(a) = x. This property allows us to "undo" the exponential and isolate x. So, let's apply the natural logarithm to both sides of our equations and see what we get. We're almost there!
Using Natural Logarithms
To solve the equations e^x = 2 and e^x = 3, we'll use the natural logarithm (ln), which is the inverse function of e^x. This is a crucial step in solving exponential equations because it allows us to isolate the variable x. Remember, the natural logarithm is simply the logarithm to the base e. Applying the natural logarithm to both sides of an equation preserves the equality, so we can confidently use this operation to solve for x.
For the first equation, e^x = 2, we take the natural logarithm of both sides: ln(e^x) = ln(2). Now, a key property of logarithms is that ln(e^x) = x. This is because the natural logarithm and the exponential function are inverses of each other. So, we get x = ln(2). That's one solution! For the second equation, e^x = 3, we do the same thing: ln(e^x) = ln(3). Again, using the property ln(e^x) = x, we find x = ln(3). And that's our second solution! We've successfully used natural logarithms to find the values of x that satisfy our equations. This technique is fundamental in solving many types of exponential equations, so it's a great tool to have in your mathematical toolkit. Now, let's summarize our results and present the final answer.
The Solutions
Alright, let's bring it all together! We've gone through quite a journey, from the initial exponential equation to the final solutions in terms of natural logarithms. Remember, we started with e^(2x) - 5e^x + 6 = 0. We used a clever substitution, y = e^x, to transform it into the quadratic equation y^2 - 5y + 6 = 0. We then solved for y by factoring, finding that y = 2 or y = 3.
Next, we substituted back e^x for y, giving us two equations: e^x = 2 and e^x = 3. Finally, we used the natural logarithm to solve for x in each equation. For e^x = 2, we found x = ln(2), and for e^x = 3, we found x = ln(3). So, the solutions to our original equation are x = ln(2) and x = ln(3). These are the exact solutions, expressed in terms of natural logarithms. Remember, natural logarithms are a powerful tool for solving exponential equations, and this problem demonstrates how useful they can be. We've successfully navigated through this problem, step by step, and arrived at the final answer. Great job, guys!
Final Answer
So, after all that awesome math work, we've arrived at our final answer! The solutions to the equation e^(2x) - 5e^x + 6 = 0, expressed in terms of natural logarithms, are:
- x = ln(2)
- x = ln(3)
And there you have it! We took a potentially tricky exponential equation and, by using a smart substitution and the magic of natural logarithms, we solved it. This is a perfect example of how breaking down a problem into smaller, manageable steps can make even the most intimidating equations solvable. Remember, the key to success in math is often recognizing patterns, using the right tools (like substitution and logarithms), and taking things one step at a time. So, keep practicing, and you'll be solving equations like this in no time! You guys rock!
