Solve Exponential Equation: Find X

Hey math enthusiasts! Today, we're diving into the world of exponential equations. Specifically, we're going to solve for x in the equation $2e^{2x} - 11e^x + 5 = 0$. This might look a bit intimidating at first, but trust me, it's totally manageable. We'll break it down step-by-step, making sure you grasp every concept along the way. Get ready to flex those mathematical muscles! We'll find all the solutions, rounding to the nearest hundredth if needed, and if there's more than one, we'll separate them with commas. Let's get started!

Understanding the Problem and Initial Steps

Alright, so we have this equation: $2e^{2x} - 11e^x + 5 = 0$. Our mission? To isolate x and find its value(s). The presence of $e^x$ and $e^{2x}$ might ring a bell – it's a quadratic equation in disguise! Because, you know, $e^{2x}$ is the same as $(e^x)^2$. This realization is the key to unlocking the solution. The first step involves recognizing this quadratic form. Then, we are going to use a substitution to simplify the equation and make it easier to work with. Let's make the substitution $u = e^x$. This transforms our equation into something much more familiar. You should be able to look at it and immediately know what to do next. It is also important to remember the properties of exponents. This step simplifies the appearance of the equation, making it much easier to solve. We're essentially trading the exponential terms for a simpler variable, making it look like a standard quadratic equation. Also, always remember the initial equation, because once we find the possible values for u, we'll have to return to the original substitution to find the values of x. The substitution is a handy trick in mathematics! It transforms complex-looking expressions into more approachable forms, helping us to identify the underlying structures and solve the problem effectively. And that's exactly what we're going to do. Let's move on to the next step, where we'll implement this substitution and simplify the equation.

Making the Substitution and Solving the Quadratic Equation

Now, let's put our plan into action. We'll substitute $u = e^x$ into the equation $2e^{2x} - 11e^x + 5 = 0$. Remember that $e^{2x}$ is the same as $(e^x)^2$, so we can rewrite the equation as: $2u^2 - 11u + 5 = 0$. See? Doesn't that look way more friendly? Now we have a standard quadratic equation in terms of u. This is where your algebra skills come in handy! To solve this quadratic equation, we can use a few different methods. Factoring is often the quickest if the equation allows it. Otherwise, we can use the quadratic formula. Let's see if we can factor this one. We are looking for two numbers that multiply to give $2 * 5 = 10$ and add up to $-11$. The numbers $-10$ and $-1$ fit the bill. So, we can rewrite the middle term of the equation as $-10u - u$. This gives us $2u^2 - 10u - u + 5 = 0$. Now, factor by grouping: $2u(u - 5) - 1(u - 5) = 0$. This simplifies to $(2u - 1)(u - 5) = 0$. We now have two possible solutions for u: either $2u - 1 = 0$ or $u - 5 = 0$. Solving these gives us u = rac{1}{2} or $u = 5$. Awesome! We've found the values of u. But remember, our goal is to find x. Now, let's put these values back into our original substitution.

Returning to the Original Variable and Finding the Solutions for x

We've successfully solved for u! Now, it's time to remember our original goal: finding the values of x. Recall that we made the substitution $u = e^x$. We found that $u$ can be equal to rac{1}{2} or $5$. Let's take each of these solutions and work backward to find the corresponding values of x. First, let's consider the case where u = rac{1}{2}. We have e^x = rac{1}{2}. To solve for x, we need to take the natural logarithm (ln) of both sides. This gives us: x = ln(rac{1}{2}). Using a calculator, we find that $x hickapprox -0.69$. Now, let's consider the second case where $u = 5$. We have $e^x = 5$. Again, we take the natural logarithm of both sides: $x = ln(5)$. Using a calculator, we find that $x hickapprox 1.61$. Therefore, we have two possible solutions for x: approximately -0.69 and 1.61. We've conquered the exponential equation! We started with a seemingly complex equation, simplified it using substitution and factoring, solved for our intermediate variable, and then returned to our original variable to find the solutions. It's a testament to how different mathematical techniques work together to solve problems. Remember that exponential functions are used in various fields, from finance to physics, so understanding how to solve these equations is a valuable skill.

Summary of Steps and Final Answer

Alright, let's recap the entire process to make sure everything's crystal clear.

1. Recognized the Form: We identified that the given equation, $2e^{2x} - 11e^x + 5 = 0$, could be treated as a quadratic equation in terms of $e^x$.
2. Made a Substitution: We simplified the equation by substituting $u = e^x$, transforming it into $2u^2 - 11u + 5 = 0$.
3. Solved the Quadratic Equation: We factored the quadratic equation to get $(2u - 1)(u - 5) = 0$, and found the solutions for u to be u = rac{1}{2} and $u = 5$.
4. Returned to the Original Variable: We substituted back $e^x$ for u and solved for x using the natural logarithm.
5. Found the Solutions: We found the two solutions for x to be approximately -0.69 and 1.61.

So, the final answer is -0.69, 1.61. We've successfully navigated the challenges of the exponential equation and found our solutions. Congratulations, you did it!