Solving Simultaneous Equations With Logarithms And Exponents

Introduction

Hey guys! Today, we're diving into a super interesting math problem: solving simultaneous equations that mix logarithms and exponents. These types of problems can seem a bit intimidating at first, but don't worry! We're going to break it down step-by-step, so it'll all make sense. We'll tackle the problem by using the properties of logarithms and exponents to simplify the equations and then solve for our variables. So, grab your calculators, and let's get started!

The ability to solve simultaneous equations is a fundamental skill in mathematics, essential not only for academic pursuits but also for various real-world applications. Simultaneous equations, which are sets of equations with multiple variables, appear frequently in fields like physics, engineering, economics, and computer science. Mastering the techniques to solve these equations opens doors to modeling and solving complex problems in these areas. In this article, we’ll focus on a specific type of simultaneous equations that involve logarithms and exponents. These equations often require a deeper understanding of logarithmic and exponential properties to simplify and find solutions. Logarithms and exponents are inverse operations, and understanding their relationship is crucial for manipulating and solving equations where they appear. The properties of logarithms, such as the product rule, quotient rule, and power rule, allow us to combine or separate logarithmic terms, making the equations easier to handle. Similarly, the properties of exponents, such as the product of powers, quotient of powers, and power of a power, help us to simplify exponential expressions. Solving these types of equations can be challenging, but with a systematic approach and a good grasp of the underlying principles, it becomes a manageable task. By working through the steps, you'll gain a better understanding of how to manipulate logarithmic and exponential expressions, a skill that is valuable in many areas of mathematics and beyond. So, let’s jump into the problem and explore how to tackle these fascinating equations!

Problem Statement

Okay, so here’s the problem we’re going to solve today. We have two equations:

1. ${2 \log _3 y = \log _5 125 + \log _3 x}$
2. ${2^y = 4^x}$

Our mission, should we choose to accept it, is to find the values of ${x}$ and ${y}$ that satisfy both of these equations. Sounds like fun, right? The first equation involves logarithms with different bases, which means we’ll need to use some logarithmic properties to simplify it. The second equation involves exponents, and we can simplify it by expressing both sides with the same base. By simplifying both equations, we’ll be able to relate ${x}$ and ${y}$, and then use substitution or another method to find the values that satisfy both equations. Remember, the key to solving these types of problems is to break them down into smaller, more manageable steps. So, let’s take a closer look at each equation and see how we can simplify them.

Step 1: Simplify the First Equation

Let’s tackle the first equation: ${2 \log _3 y = \log _5 125 + \log _3 x}$. We need to simplify this bad boy. First up, we can simplify ${\log _5 125}$. Think about it: 5 to what power gives us 125? That's right, it's 3! So, ${\log _5 125 = 3}$. Now, our equation looks like this: ${2 \log _3 y = 3 + \log _3 x}$. The next step is to use the power rule of logarithms, which states that ${a \log _b c = \log _b c^a}$. Applying this rule to the term ${2 \log _3 y}$, we get ${\log _3 y^2}$. So, our equation now becomes ${\log _3 y^2 = 3 + \log _3 x}$. To further simplify, we need to get rid of the constant term 3. We can rewrite 3 as a logarithm with base 3. Since ${3 = \log _3 3^3 = \log _3 27}$, we can substitute 3 with ${\log _3 27}$. Now, our equation looks like this: ${\log _3 y^2 = \log _3 27 + \log _3 x}$. Next, we can use the product rule of logarithms, which states that ${\log _b m + \log _b n = \log _b (mn)}$. Applying this rule to the right side of the equation, we get ${\log _3 27x}$. So, now we have ${\log _3 y^2 = \log _3 27x}$. Since the logarithms on both sides have the same base, we can equate the arguments, giving us ${y^2 = 27x}$. This is a much simpler equation than what we started with, and it relates ${x}$ and ${y}$. We’ve made significant progress in simplifying the first equation! This simplified equation will be super helpful when we combine it with the second equation to solve for ${x}$ and ${y}$. So, let’s keep this result in mind and move on to simplifying the second equation.

In this step, we utilized several key properties of logarithms to simplify the first equation. The ability to simplify logarithmic expressions is a crucial skill in mathematics, and understanding these properties is essential for solving more complex problems. We started by simplifying ${\log _5 125}$, which is a straightforward application of the definition of logarithms. Then, we used the power rule to rewrite ${2 \log _3 y}$ as ${\log _3 y^2}$. This step is crucial because it allows us to combine logarithmic terms more easily. Next, we rewrote the constant 3 as a logarithm with base 3, which is a common technique when dealing with logarithmic equations. This allowed us to express all terms in the equation as logarithms with the same base. Finally, we used the product rule of logarithms to combine the terms on the right side of the equation. By equating the arguments of the logarithms, we obtained a simple algebraic equation relating ${x}$ and ${y}$. This is a significant step towards solving the system of equations. Simplifying equations is often the first step in solving mathematical problems, and it's important to develop a systematic approach to this process. By breaking down the problem into smaller steps and applying the appropriate properties, we can transform complex equations into simpler, more manageable forms.

Step 2: Simplify the Second Equation

Alright, let's move on to the second equation: ${2^y = 4^x}$. This one looks a bit simpler, but we still need to do some work. The key here is to recognize that we can express both sides of the equation with the same base. We know that ${4 = 2^2}$, so we can rewrite ${4^x}$ as ${(2^2)^x}$. Using the power of a power rule, which states that ${(a^m)^n = a^{mn}}$, we can simplify ${(2^2)^x}$ to ${2^{2x}}$. Now, our equation looks like this: ${2^y = 2^{2x}}$. Since the bases are the same, we can equate the exponents, giving us ${y = 2x}$. Boom! That was much easier, wasn't it? We've now simplified the second equation to a very straightforward relationship between ${x}$ and ${y}$. This simple equation will be extremely helpful when we combine it with the simplified form of the first equation. We’re making great progress in solving this system of equations. By simplifying both equations, we’ve made the problem much more manageable.

In this step, we focused on simplifying the second equation, which involves exponents. The key to simplifying exponential equations is to express all terms with the same base. By recognizing that 4 can be written as ${2^2}$, we were able to rewrite the equation in terms of base 2. This is a common strategy when dealing with exponential equations, and it often simplifies the problem significantly. The power of a power rule, ${(a^m)^n = a^{mn}}$, is a fundamental property of exponents that allows us to simplify expressions involving exponents raised to another exponent. Applying this rule, we transformed ${(2^2)^x}$ into ${2^{2x}}$. This step is crucial because it allows us to equate the exponents. Once we had both sides of the equation expressed with the same base, we could simply equate the exponents to obtain a linear equation relating ${x}$ and ${y}$. This resulted in the equation ${y = 2x}$, which is a very simple relationship between the two variables. Simplifying exponential equations often involves identifying common bases and using the properties of exponents to rewrite the equation in a more manageable form. This step demonstrates the power of this approach and how it can lead to a straightforward solution. By simplifying the second equation, we have obtained another key piece of information that will help us solve the system of equations. Now, we have two simple equations relating ${x}$ and ${y}$, which we can use to find the values that satisfy both equations.

Step 3: Solve the Simultaneous Equations

Okay, now for the grand finale! We have two simplified equations:

1. ${y^2 = 27x}$
2. ${y = 2x}$

We can use substitution to solve for ${x}$ and ${y}$. Since we have ${y = 2x}$, we can substitute ${2x}$ for ${y}$ in the first equation. This gives us ${(2x)^2 = 27x}$. Expanding ${(2x)^2}$, we get ${4x^2 = 27x}$. To solve for ${x}$, we need to rearrange the equation into a quadratic equation. Subtracting ${27x}$ from both sides, we get ${4x^2 - 27x = 0}$. Now, we can factor out an ${x}$, giving us ${x(4x - 27) = 0}$. This equation has two possible solutions for ${x}$: either ${x = 0}$ or ${4x - 27 = 0}$. If ${4x - 27 = 0}$, then ${4x = 27}$, and ${x = \frac{27}{4}}$. So, we have two possible values for ${x}$: ${x = 0}$ or ${x = \frac{27}{4}}$. Now, we need to find the corresponding values for ${y}$. We can use the equation ${y = 2x}$ to do this. If ${x = 0}$, then ${y = 2(0) = 0}$. If ${x = \frac{27}{4}}$, then ${y = 2(\frac{27}{4}) = \frac{27}{2}}$. So, we have two possible solutions for ${(x, y)}$: ${(0, 0)}$ and ${(\frac{27}{4}, \frac{27}{2})}$. However, we need to check if these solutions are valid in the original equations, especially the logarithmic equation, because logarithms are not defined for non-positive arguments. Let's check the first solution, ${(0, 0)}$. Plugging these values into the first original equation, ${2 \log _3 y = \log _5 125 + \log _3 x}$, we get ${2 \log _3 0 = \log _5 125 + \log _3 0}$. Since ${\log _3 0}$ is undefined, the solution ${(0, 0)}$ is not valid. Now, let's check the second solution, ${(\frac{27}{4}, \frac{27}{2})}$. Plugging these values into the original equations: For the first equation, ${2 \log _3 y = \log _5 125 + \log _3 x}$, we have ${2 \log _3 (\frac{27}{2}) = \log _5 125 + \log _3 (\frac{27}{4})}$. Simplifying, ${2 \log _3 (\frac{27}{2}) = 3 + \log _3 (\frac{27}{4})}$. For the second equation, ${2^y = 4^x}$, we have ${2^{\frac{27}{2}} = 4^{\frac{27}{4}}}$ which is equivalent to {2^{\frac{27}{2}} = (2^2)^{\frac{27}{4}} = 2^{\frac{27}{2}}\, so it is true. The first equation also holds true after evaluating both sides. Therefore, the only valid solution is \((\frac{27}{4}, \frac{27}{2})}. We've successfully solved the simultaneous equations! We found the values of ${x}$ and ${y}$ that satisfy both equations by using substitution and checking for validity. This was a challenging problem, but by breaking it down into steps and using the properties of logarithms and exponents, we were able to find the solution. Great job, everyone!

In this final step, we combined the simplified equations to solve for ${x}$ and ${y}$. We used the method of substitution, which is a common technique for solving simultaneous equations. By substituting ${y = 2x}$ into the equation ${y^2 = 27x}$, we obtained a quadratic equation in terms of ${x}$. Solving quadratic equations is a fundamental skill in algebra, and there are several methods we can use, such as factoring, completing the square, or using the quadratic formula. In this case, we were able to factor the quadratic equation, which made it easy to find the solutions for ${x}$. Once we had the values of ${x}$, we could use the equation ${y = 2x}$ to find the corresponding values of ${y}$. This gave us two possible solutions for ${(x, y)}$. However, it's crucial to check the validity of the solutions, especially when dealing with logarithmic equations. Logarithms are only defined for positive arguments, so we need to make sure that the solutions we obtain do not result in taking the logarithm of a non-positive number. By checking the solutions in the original equations, we found that one of the solutions was not valid because it involved taking the logarithm of zero. The other solution, ${(\frac{27}{4}, \frac{27}{2})}$, was valid and satisfied both original equations. Checking the validity of solutions is an important step in solving mathematical problems, as it ensures that the solutions we obtain are meaningful in the context of the problem. This step demonstrates the importance of being thorough and careful when solving mathematical equations. By combining the simplified equations, solving for the variables, and checking the validity of the solutions, we were able to successfully solve the simultaneous equations.

Conclusion

And there you have it! We've successfully solved a system of simultaneous equations involving logarithms and exponents. It might have seemed daunting at first, but by breaking it down into smaller steps, using the properties of logarithms and exponents, and carefully checking our solutions, we were able to find the answer. Remember, the key to tackling these types of problems is to stay organized, understand the properties of the functions involved, and practice, practice, practice! Keep up the great work, guys, and you'll be solving these equations like a pro in no time!

Solving simultaneous equations that involve logarithms and exponents can be a challenging but rewarding mathematical exercise. By simplifying the equations using properties of logarithms and exponents, we can transform them into a more manageable form. Substitution and other algebraic techniques can then be used to solve for the variables. Checking the validity of solutions is crucial, especially with logarithms, to ensure that the solutions are meaningful in the context of the original problem. Mastering these techniques provides a solid foundation for tackling more complex mathematical problems in various fields. The ability to solve such equations is not only a valuable skill in mathematics but also in many areas of science and engineering where mathematical modeling is used to solve real-world problems. The process of solving these equations reinforces the importance of a systematic approach to problem-solving, breaking down complex problems into smaller, more manageable steps, and applying the appropriate tools and techniques. So, keep practicing and exploring, and you’ll become more confident and proficient in solving these types of equations.