Solving Exponential Equations: A Step-by-Step Guide

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Hey guys! Today, we're diving into the world of exponential equations and tackling a common problem type you might encounter in your math journey. Specifically, we'll break down the solution to the equation 5e2xβˆ’4=115e^{2x} - 4 = 11. Don't worry, it's not as intimidating as it looks! We'll take it step by step, so you'll be solving these like a pro in no time. So, grab your thinking caps, and let's get started!

Understanding Exponential Equations

Before we jump into the solution, let's quickly recap what exponential equations are all about. Exponential equations are equations where the variable appears in the exponent. These types of equations often describe phenomena that grow or decay exponentially, such as population growth, radioactive decay, or compound interest. Recognizing the structure of an exponential equation is the first step in solving it. Our equation, 5e2xβˆ’4=115e^{2x} - 4 = 11, clearly fits this description because the variable 'x' is nestled up there in the exponent of the exponential term, e2xe^{2x}. The key to solving these equations lies in isolating the exponential term and then using logarithms to "undo" the exponentiation. Think of it like this: exponentiation and logarithms are mathematical opposites, like addition and subtraction or multiplication and division. By understanding this fundamental relationship, you'll be well-equipped to tackle a wide range of exponential equations. We'll see exactly how this works as we solve our example problem. Remember, the goal is to get the exponential term by itself, and then use the properties of logarithms to bring the variable down from the exponent.

Step-by-Step Solution for 5e2xβˆ’4=115e^{2x} - 4 = 11

Alright, let's get down to business and solve this equation! Remember, our goal is to isolate the exponential term and then use logarithms. Here’s a breakdown of the steps:

1. Isolate the Exponential Term

Our first mission is to get the term with the exponent, which is 5e2x5e^{2x}, all by itself on one side of the equation. To do this, we need to get rid of that pesky -4. We can do this by adding 4 to both sides of the equation. This is a fundamental principle of equation solving: whatever you do to one side, you must do to the other to maintain the balance. So, we have:

5e2xβˆ’4+4=11+45e^{2x} - 4 + 4 = 11 + 4

This simplifies to:

5e2x=155e^{2x} = 15

Great! We're one step closer. Now, we still have that 5 multiplying the exponential term. To get rid of it, we'll divide both sides of the equation by 5:

rac{5e^{2x}}{5} = rac{15}{5}

This simplifies to:

e2x=3e^{2x} = 3

Awesome! We've successfully isolated the exponential term. Now the real fun begins.

2. Apply the Natural Logarithm

This is where logarithms come into play. Remember, logarithms are the inverse operation of exponentiation. Since our exponential term has a base of 'e' (Euler's number, approximately 2.718), we're going to use the natural logarithm, denoted as "ln". The natural logarithm is simply the logarithm to the base 'e'. The key property of logarithms we'll use here is that ln⁑(ex)=x\ln(e^x) = x. In other words, the natural logarithm "undoes" the exponential function with base 'e'. To apply the natural logarithm, we take the natural log of both sides of the equation:

ln⁑(e2x)=ln⁑(3)\ln(e^{2x}) = \ln(3)

Using the property we just mentioned, the left side simplifies beautifully:

2x=ln⁑(3)2x = \ln(3)

See how the exponent came down? That's the magic of logarithms!

3. Solve for x

We're in the home stretch now! We have 2x=ln⁑(3)2x = \ln(3). To isolate 'x', we simply divide both sides of the equation by 2:

2x2=ln⁑(3)2\frac{2x}{2} = \frac{\ln(3)}{2}

This gives us our solution:

x=ln⁑(3)2x = \frac{\ln(3)}{2}

And there you have it! We've successfully solved the exponential equation. The solution is x=ln⁑(3)2x = \frac{\ln(3)}{2}. This corresponds to option C in the original problem.

Why This Solution Makes Sense

It's always a good idea to take a moment and think about why our solution makes sense. We started with the equation 5e2xβˆ’4=115e^{2x} - 4 = 11. We found that x=ln⁑(3)2x = \frac{\ln(3)}{2}. Let's plug this back into the original equation and see if it works:

5e2(ln⁑(3)2)βˆ’4=5eln⁑(3)βˆ’45e^{2(\frac{\ln(3)}{2})} - 4 = 5e^{\ln(3)} - 4

Remember that eln⁑(x)=xe^{\ln(x)} = x, so:

5eln⁑(3)βˆ’4=5(3)βˆ’4=15βˆ’4=115e^{\ln(3)} - 4 = 5(3) - 4 = 15 - 4 = 11

It works! This confirms that our solution is correct. This process of checking your answer is a valuable habit to develop, especially in math. It helps you catch any potential errors and build confidence in your solutions.

Common Mistakes to Avoid

When solving exponential equations, there are a few common pitfalls that students often stumble into. Being aware of these mistakes can help you avoid them.

  • Forgetting to Isolate the Exponential Term First: This is perhaps the most common mistake. You can't apply logarithms until you've isolated the exponential term. Make sure to perform any necessary additions, subtractions, multiplications, or divisions to get the exponential term by itself on one side of the equation.
  • Incorrectly Applying Logarithm Properties: Logarithms have specific properties that you need to follow carefully. For example, ln⁑(aβˆ—b)\ln(a*b) is not the same as ln⁑(a)βˆ—ln⁑(b)\ln(a) * \ln(b). Remember the key properties, such as ln⁑(ex)=x\ln(e^x) = x and eln⁑(x)=xe^{\ln(x)} = x, and use them correctly.
  • Dividing Before Taking the Logarithm: You might be tempted to divide by the coefficient of the exponential term after taking the logarithm. However, you need to isolate the exponential term before applying the logarithm. This ensures you're applying the logarithm to the correct expression.
  • Rounding Too Early: If you're using a calculator to find the decimal approximation of the natural logarithm, avoid rounding the result until the very end of the problem. Rounding too early can introduce inaccuracies into your final answer.

By being mindful of these common mistakes, you can increase your accuracy and confidence in solving exponential equations.

Practice Makes Perfect

The best way to master solving exponential equations is to practice, practice, practice! Try working through similar problems with different numbers and variations. You can find plenty of practice problems in your textbook, online resources, or from your teacher. The more you practice, the more comfortable you'll become with the steps involved and the more easily you'll be able to recognize and solve these types of equations. Don't be afraid to make mistakes – they're a valuable part of the learning process. Just make sure to learn from your mistakes and keep pushing forward!

Conclusion

So, guys, we've successfully navigated the world of exponential equations and found the solution to 5e2xβˆ’4=115e^{2x} - 4 = 11, which is x=ln⁑(3)2x = \frac{\ln(3)}{2}. We covered the importance of isolating the exponential term, using the natural logarithm, and avoiding common mistakes. Remember, solving exponential equations is a fundamental skill in mathematics, and with practice, you'll become a pro in no time. Keep practicing, keep learning, and keep conquering those math challenges! You've got this!