Solve Exponential Equations With Natural Logarithms

Introduction

Hey guys! Today, we're diving into the fascinating world of exponential equations, and we're going to tackle a specific problem using our trusty friend, the natural logarithm. Exponential equations might seem a bit intimidating at first, but don't worry, we'll break it down step by step so you can conquer them with confidence. Our main goal here is to solve an equation that looks like this: $e^{2x} - 3e^x - 18 = 0$. Sounds fun, right? We'll not only find the exact values using common logarithms but also approximate solutions to four decimal places. So, buckle up and let's get started!

Understanding Exponential Equations

Before we jump into solving the equation, let's make sure we're all on the same page about what exponential equations are. An exponential equation is simply an equation where the variable appears in the exponent. Think of it like this: instead of having $x^2$, you have something like $2^x$. The key here is that the variable is up in the exponent, making things a little more interesting.

Now, why are these equations important? Well, exponential functions pop up all over the place in the real world. They're used to model things like population growth, radioactive decay, compound interest, and even the spread of diseases. So, understanding how to solve exponential equations is not just a cool math trick—it's a valuable skill for understanding the world around us. To solve these equations, we often need to use logarithms, which are essentially the “opposite” of exponentials. Logarithms allow us to bring the variable down from the exponent, making the equation easier to handle. In our case, we'll be using the natural logarithm, which has a special base called e (approximately 2.71828). The natural logarithm is written as “ln” and is super handy for equations involving the exponential function with base e.

Setting Up the Equation

Okay, let's get back to our equation: $e^{2x} - 3e^x - 18 = 0$. At first glance, it might not look like your typical quadratic equation, but here's a neat trick. We can use a substitution to make it look more familiar. Let's set $y = e^x$. This is a common technique when dealing with exponential equations that have a certain structure. By making this substitution, we're essentially simplifying the equation into a form we already know how to solve.

Now, if $y = e^x$, then $y^2$ would be $(e^x)^2$, which is the same as $e^{2x}$. So, we can rewrite our original equation using this new variable y. Instead of $e^{2x} - 3e^x - 18 = 0$, we now have $y^2 - 3y - 18 = 0$. See? It looks much more like a quadratic equation now! This is a crucial step because we can use all our knowledge of solving quadratic equations to find the values of y. Once we have the values for y, we can then substitute back to find the values for x. This substitution technique is a powerful tool in solving many types of exponential equations, and it's something you'll use again and again as you tackle more complex problems.

Solving the Quadratic Equation

Great! Now that we've transformed our exponential equation into a quadratic equation, $y^2 - 3y - 18 = 0$, we can solve it using methods you probably already know. There are several ways to solve quadratic equations, such as factoring, using the quadratic formula, or completing the square. For this particular equation, factoring is the most straightforward method. Factoring involves finding two numbers that multiply to the constant term (-18) and add up to the coefficient of the linear term (-3). In this case, those numbers are -6 and +3.

So, we can rewrite the quadratic equation as $(y - 6)(y + 3) = 0$. This means that either $(y - 6) = 0$ or $(y + 3) = 0$. Solving these two simple equations gives us the solutions for y: $y = 6$ or $y = -3$. These are the values of y that make our quadratic equation true. But remember, we're not trying to solve for y; we want to find the values of x that satisfy our original exponential equation. So, we need to substitute back and use the fact that $y = e^x$. This is where the magic of logarithms comes into play. We'll take these y values and use them to find the corresponding x values in the next step.

Back-Substitution and Natural Logarithms

Alright, we've found our y values: $y = 6$ and $y = -3$. Now, it's time to substitute back and find the x values. Remember, we said $y = e^x$, so we now have two equations to solve: $e^x = 6$ and $e^x = -3$. This is where the natural logarithm comes in handy. The natural logarithm (ln) is the inverse function of the exponential function with base e. In other words, if $e^x = a$, then $x = ln(a)$.

Let's start with $e^x = 6$. To solve for x, we take the natural logarithm of both sides: $ln(e^x) = ln(6)$. Since $ln(e^x)$ simplifies to x, we have $x = ln(6)$. This is one of our solutions, and it's an exact value. We can use a calculator to find an approximate value to four decimal places, which is approximately 1.7918.

Now, let's look at the second equation: $e^x = -3$. Here's a crucial point: the exponential function $e^x$ is always positive for any real number x. It can never be negative or zero. This is because e raised to any power will always result in a positive number. Therefore, the equation $e^x = -3$ has no real solutions. So, we can disregard this case. This is an important thing to remember when solving exponential equations: always check if your solutions make sense in the context of the original equation.

Exact and Approximate Solutions

So, we've found that $x = ln(6)$ is the only solution to our original exponential equation $e^{2x} - 3e^x - 18 = 0$. This is the exact value of the solution, expressed using the natural logarithm. Now, let's talk about approximate solutions. While exact values are great, sometimes we need a decimal approximation to get a better sense of the number. To find the approximate solution, we can use a calculator to evaluate $ln(6)$.

When you plug $ln(6)$ into a calculator, you get approximately 1.791759469. But we were asked to round our solution to four decimal places. So, we look at the fifth decimal place (which is 5) to decide whether to round up or down. Since it's 5 or greater, we round up the fourth decimal place. Therefore, the approximate solution to four decimal places is 1.7918. It’s always a good practice to specify both the exact solution (using logarithms) and the approximate solution (as a decimal) when solving exponential equations. This gives a complete picture of the solution set.

Solution Set

Okay, guys, let's wrap things up by writing out the solution set. The solution set is simply the set of all values of x that satisfy the original equation. In our case, we found one real solution: $x = ln(6)$. So, the solution set is simply {$ln(6)$}. This is the exact solution, and we can also express it as an approximate solution, which we found to be 1.7918 when rounded to four decimal places.

So, to summarize, we started with the exponential equation $e^{2x} - 3e^x - 18 = 0$, used a substitution to transform it into a quadratic equation, solved the quadratic equation, and then used natural logarithms to find the value of x. We expressed our solution both as an exact value using the natural logarithm and as an approximate value to four decimal places. This is a standard approach to solving exponential equations, and you'll find that this method works for many similar problems.

Conclusion

And there you have it! We've successfully solved the exponential equation $e^{2x} - 3e^x - 18 = 0$. We found the exact solution, $x = ln(6)$, and the approximate solution, $x ≈ 1.7918$. Remember, the key to solving these types of equations is to use substitution to simplify them and then apply logarithms to isolate the variable. Exponential equations might seem tricky at first, but with practice, you'll become a pro at solving them. Keep practicing, and you'll master these equations in no time. Keep up the great work, and I'll see you in the next math adventure!