Solving Exponential Equations 5e^{7x}=1055 With Logarithms
Introduction
Hey guys! Today, we're diving into the exciting world of exponential equations and how to solve them like pros. Exponential equations might seem intimidating at first, but with the right approach and a solid understanding of logarithms, you'll be cracking these problems in no time. We're going to tackle a specific example, step-by-step, making sure everything is crystal clear. So, buckle up and let's get started!
The Exponential Equation
Our mission today is to solve the following exponential equation:
5e^{7x} = 1055
This equation involves an exponential term, e^{7x}, where 'e' is the base of the natural logarithm (approximately 2.71828). Our goal is to isolate 'x', and to do that, we'll be using the power of logarithms. Specifically, we'll be expressing our solution in terms of natural logarithms (ln) and understand how we could also use common logarithms (log base 10) if we wanted to. Understanding how to manipulate and solve these equations is super important in many fields, from finance to science, so let's get this down!
Isolating the Exponential Term
The first crucial step in solving any exponential equation is to isolate the exponential term. In our case, that's e^{7x}. We need to get this term all by itself on one side of the equation. Currently, it's being multiplied by 5. No sweat! We can easily undo this multiplication by dividing both sides of the equation by 5. This maintains the balance of the equation, which is super important in algebra. Remember, whatever you do to one side, you have to do to the other!
So, let’s do it:
5e^{7x} / 5 = 1055 / 5
This simplifies beautifully to:
e^{7x} = 211
Awesome! We've successfully isolated the exponential term. Now we’re one step closer to solving for 'x'. This step is crucial because it sets us up to use logarithms effectively. We can't just jump straight to solving for 'x' when the exponential term is tangled up with other numbers. Think of it like untangling your headphones before you can listen to music – gotta get the basics sorted first!
Applying the Natural Logarithm
Now for the magic trick! To get 'x' out of the exponent, we need to use the inverse operation of exponentiation, which is the logarithm. Since our exponential term has a base of 'e', the natural logarithm (ln) is the perfect tool for the job. The natural logarithm is simply the logarithm with base 'e'. Remember that logarithms basically ask the question: "To what power must I raise the base to get this number?"
We'll apply the natural logarithm to both sides of our equation. Again, it’s all about balance – what we do to one side, we must do to the other:
ln(e^{7x}) = ln(211)
Here’s where the magic really happens. One of the key properties of logarithms is that ln(a^b) = b * ln(a). In our case, this means ln(e^{7x}) simplifies to 7x * ln(e). But wait, there's more! The natural logarithm of 'e', ln(e), is equal to 1. This is because 'e' raised to the power of 1 is 'e'. So, 7x * ln(e) becomes simply 7x.
Our equation now looks much cleaner:
7x = ln(211)
We’ve successfully used the natural logarithm to bring 'x' down from the exponent and into a place where we can actually solve for it. This step is a classic example of how logarithms are used to unravel exponential equations. It's like using a special key to unlock a secret – in this case, the secret is the value of 'x'!
Solving for x
We're almost there! We have the equation 7x = ln(211). To finally isolate 'x', we need to get rid of the 7 that's multiplying it. Just like we used division to undo multiplication earlier, we'll divide both sides of the equation by 7:
7x / 7 = ln(211) / 7
This simplifies to:
x = ln(211) / 7
Boom! We've done it. We've solved for 'x'. Our solution is x = ln(211) / 7. This is the exact solution, expressed in terms of the natural logarithm. It's a perfectly valid answer, and in many cases, it's the preferred way to express the solution, especially when you need high accuracy.
If you need a decimal approximation, you can plug ln(211) / 7 into a calculator. You'll find that it's approximately equal to 0.759. But remember, the exact solution ln(211) / 7 is the most precise answer. This final step is like putting the last piece of the puzzle in place. We started with a complex-looking equation, and now we have a clear, concise solution for 'x'.
Expressing the Solution Set
In mathematics, the solution set is the set of all values that satisfy an equation. In our case, we have only one solution, which is x = ln(211) / 7. So, we can express the solution set as:
{ln(211) / 7}
This notation simply means that the set contains the single value ln(211) / 7. It's a formal way of stating our answer, making it clear that we've found all the values of 'x' that make the original equation true. Think of it as putting a neat little box around our answer, so it’s clear and easy to identify.
Using Common Logarithms
Now, let's talk about common logarithms. While we've used natural logarithms to solve this equation, we could have used common logarithms (log base 10) as well. The process is very similar, but let's walk through it to see the parallels and differences. It's always good to have options in your mathematical toolkit!
Applying the Common Logarithm
Instead of using the natural logarithm, we'll apply the common logarithm (log base 10) to both sides of our equation e^{7x} = 211:
log(e^{7x}) = log(211)
Now, we use the same logarithm property as before: log(a^b) = b * log(a). This gives us:
7x * log(e) = log(211)
Notice that log(e) is not equal to 1 like ln(e) was. log(e) is approximately 0.4343. This is because we're asking, “To what power must we raise 10 to get e?” It’s a different question than asking, “To what power must we raise e to get e?”
Solving for x with Common Logarithms
To solve for 'x', we need to divide both sides of the equation by 7 * log(e):
x = log(211) / (7 * log(e))
This is our solution expressed in terms of common logarithms. It looks a bit different from our solution using natural logarithms, but it represents the same value. To see this, you could plug both ln(211) / 7 and log(211) / (7 * log(e)) into a calculator and you'll get the same decimal approximation (approximately 0.759). This is a cool demonstration of how different logarithmic bases can be used to solve the same problem!
Change of Base Formula
You might be wondering how these two solutions are equivalent, given that they look so different. The secret lies in the change of base formula for logarithms. This formula allows us to convert a logarithm from one base to another. It states:
log_b(a) = log_c(a) / log_c(b)
Where log_b(a) is the logarithm of 'a' with base 'b', and log_c is a logarithm with any other base 'c'.
If we apply this formula to our situation, we can convert our common logarithm solution to a natural logarithm solution (or vice versa). This formula is a powerful tool for working with logarithms, and it helps to bridge the gap between different logarithmic bases. It's like having a universal translator for logarithms!
Solution Set in Terms of Common Logarithms
Just like before, we can express the solution set using our common logarithm solution:
{log(211) / (7 * log(e))}
This solution set contains the same value as our natural logarithm solution set, but it's expressed in a different form. Both are correct, and the choice of which one to use often depends on the context or the specific requirements of the problem.
Conclusion
Woohoo! We've successfully solved the exponential equation 5e^{7x} = 1055 and expressed the solution set in terms of both natural logarithms and common logarithms. We've seen how to isolate the exponential term, apply logarithms to bring the exponent down, and solve for 'x'. We also explored the change of base formula, which helps us understand the relationship between different logarithmic bases. Solving exponential equations is a fundamental skill in mathematics, and you guys now have the tools to tackle these problems with confidence.
Remember, the key to mastering exponential equations is practice. So, try out some more examples, and don't be afraid to experiment with different approaches. Keep practicing, and you'll become an exponential equation-solving ninja in no time! You've got this!
