Solving Logarithmic Equations: A Step-by-Step Guide

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Hey guys! Let's dive into a cool algebra problem. We're given some equations with exponents and logarithms, and our goal is to find the value of an expression. Sounds fun, right? Don't worry, we'll break it down step by step, so it's super easy to follow along. We'll start with the problem, understand what it's asking, and then use some clever tricks to solve it. Let's get started and make sure to use all the right keywords so this article ranks well!

Understanding the Problem: The Core of the Matter

Alright, so here's what we're dealing with. We have three positive real numbers, which are x, y, and z. These numbers are related by the following equations:

  • xlog⁡2(yz)=212⋅58x^{\log_{2}(yz)} = 2^{12} \cdot 5^{8}
  • ylog⁡2(zx)=26⋅52y^{\log_{2}(zx)} = 2^{6} \cdot 5^{2}
  • zlog⁡2(xy)=210⋅510z^{\log_{2}(xy)} = 2^{10} \cdot 5^{10}

Our mission, should we choose to accept it, is to find the value of the expression, or some related value! Now, at first glance, these equations might seem a bit intimidating, with those exponents and logarithms. But trust me, we can handle it. The key here is to use the properties of logarithms and exponents to simplify these equations and make them easier to solve. We'll aim to transform these equations into a more manageable form, like linear equations, which we all know and love (or at least, we're familiar with them!). This involves applying logarithmic rules to both sides of the equations, which will help us unravel the relationships between x, y, and z. Keep in mind that understanding the problem is the first and most crucial step in solving any math problem. If you get this part right, you're already halfway there! This will set the foundation for the next steps and allow you to see the solution more clearly. So, take a deep breath, and let's get into the nitty-gritty. Now, let’s go over the logarithmic equations and their implications.

Breaking Down the Logarithmic Equations: The Blueprint

Okay, let's take a closer look at these equations. Notice that each equation has a term like xlog⁡2(yz)x^{\log_{2}(yz)}. This means that x is raised to the power of a logarithm. This is where our knowledge of logarithms comes in handy. Remember the basic logarithmic identity: log⁡b(ac)=c⋅log⁡b(a)\log_b(a^c) = c \cdot \log_b(a)? We're going to use this property. Taking the base-2 logarithm of both sides of each equation will be our first move. Why base-2? Because the logarithm in the exponents is also base-2. This is not a coincidence! This allows us to simplify things neatly. Let's do it step by step to avoid any confusion. For the first equation, xlog⁡2(yz)=212⋅58x^{\log_{2}(yz)} = 2^{12} \cdot 5^{8}, we take the base-2 logarithm of both sides. This gives us: log⁡2(xlog⁡2(yz))=log⁡2(212⋅58)\log_{2}(x^{\log_{2}(yz)}) = \log_{2}(2^{12} \cdot 5^{8}). Using the logarithmic identity, we get: log⁡2(yz)⋅log⁡2(x)=log⁡2(212)+log⁡2(58)\log_{2}(yz) \cdot \log_{2}(x) = \log_{2}(2^{12}) + \log_{2}(5^{8}). And we simplify further to get log⁡2(yz)⋅log⁡2(x)=12+8⋅log⁡2(5)\log_{2}(yz) \cdot \log_{2}(x) = 12 + 8 \cdot \log_{2}(5). We will apply the same approach to the other two equations. This will give us a system of equations where the unknowns are logarithms of x, y, and z. Let’s get into applying the logarithm property to the other equations.

Applying Logarithmic Properties: The Transformation

Let’s apply the same logarithmic transformation to the remaining two equations. For the second equation, ylog⁡2(zx)=26⋅52y^{\log_{2}(zx)} = 2^{6} \cdot 5^{2}, we take the base-2 logarithm of both sides, which yields log⁡2(ylog⁡2(zx))=log⁡2(26⋅52)\log_{2}(y^{\log_{2}(zx)}) = \log_{2}(2^{6} \cdot 5^{2}). Then we can simplify to get: log⁡2(zx)⋅log⁡2(y)=log⁡2(26)+log⁡2(52)\log_{2}(zx) \cdot \log_{2}(y) = \log_{2}(2^{6}) + \log_{2}(5^{2}). Further simplifying it gives us: log⁡2(zx)⋅log⁡2(y)=6+2⋅log⁡2(5)\log_{2}(zx) \cdot \log_{2}(y) = 6 + 2 \cdot \log_{2}(5). For the third equation, zlog⁡2(xy)=210⋅510z^{\log_{2}(xy)} = 2^{10} \cdot 5^{10}, taking the base-2 logarithm on both sides results in log⁡2(zlog⁡2(xy))=log⁡2(210⋅510)\log_{2}(z^{\log_{2}(xy)}) = \log_{2}(2^{10} \cdot 5^{10}). Therefore, we simplify the expression to log⁡2(xy)⋅log⁡2(z)=log⁡2(210)+log⁡2(510)\log_{2}(xy) \cdot \log_{2}(z) = \log_{2}(2^{10}) + \log_{2}(5^{10}). Furthermore, it simplifies into log⁡2(xy)⋅log⁡2(z)=10+10⋅log⁡2(5)\log_{2}(xy) \cdot \log_{2}(z) = 10 + 10 \cdot \log_{2}(5). Now we have three equations. Each equation includes logarithms of x, y, and z. This is excellent because it means we can manipulate these equations to eliminate variables or find relationships. Now, we will simplify each logarithmic expression further using logarithmic properties and create a linear system of equations to solve for the logarithms.

Simplifying and Solving: The Core Steps

Alright, now that we've transformed our equations using logarithms, let's simplify them and solve for the unknowns. Remember, our goal is to find the value of the given expression, or related value. To do this, we'll need to solve for log⁡2(x)\log_{2}(x), log⁡2(y)\log_{2}(y), and log⁡2(z)\log_{2}(z). We already have three equations. But before we get to the system of equations, let's simplify the logarithmic expressions on the left-hand side of each equation. For the first equation, log⁡2(yz)⋅log⁡2(x)=12+8⋅log⁡2(5)\log_{2}(yz) \cdot \log_{2}(x) = 12 + 8 \cdot \log_{2}(5). Using the logarithm property log⁡(ab)=log⁡(a)+log⁡(b)\log(ab) = \log(a) + \log(b), we can rewrite log⁡2(yz)\log_{2}(yz) as log⁡2(y)+log⁡2(z)\log_{2}(y) + \log_{2}(z). Our equation becomes: (log⁡2(y)+log⁡2(z))⋅log⁡2(x)=12+8⋅log⁡2(5)(\log_{2}(y) + \log_{2}(z)) \cdot \log_{2}(x) = 12 + 8 \cdot \log_{2}(5). Similarly, for the second equation, log⁡2(zx)⋅log⁡2(y)=6+2⋅log⁡2(5)\log_{2}(zx) \cdot \log_{2}(y) = 6 + 2 \cdot \log_{2}(5), we can rewrite log⁡2(zx)\log_{2}(zx) as log⁡2(z)+log⁡2(x)\log_{2}(z) + \log_{2}(x). Now, our second equation will look like this: (log⁡2(z)+log⁡2(x))⋅log⁡2(y)=6+2⋅log⁡2(5)(\log_{2}(z) + \log_{2}(x)) \cdot \log_{2}(y) = 6 + 2 \cdot \log_{2}(5). The third equation, log⁡2(xy)⋅log⁡2(z)=10+10⋅log⁡2(5)\log_{2}(xy) \cdot \log_{2}(z) = 10 + 10 \cdot \log_{2}(5), can be simplified as follows: (log⁡2(x)+log⁡2(y))⋅log⁡2(z)=10+10⋅log⁡2(5)(\log_{2}(x) + \log_{2}(y)) \cdot \log_{2}(z) = 10 + 10 \cdot \log_{2}(5). Now we have a system of three equations that are easier to work with, making the problem simpler than it initially seemed.

Creating a Linear System: The Strategy

Now we're going to create a linear system of equations. We'll treat log⁡2(x)\log_{2}(x), log⁡2(y)\log_{2}(y), and log⁡2(z)\log_{2}(z) as our variables. Let's denote a=log⁡2(x)a = \log_{2}(x), b=log⁡2(y)b = \log_{2}(y), and c=log⁡2(z)c = \log_{2}(z). The equations become:

  1. (b+c)a=12+8⋅log⁡2(5)(b + c)a = 12 + 8 \cdot \log_{2}(5)
  2. (c+a)b=6+2⋅log⁡2(5)(c + a)b = 6 + 2 \cdot \log_{2}(5)
  3. (a+b)c=10+10⋅log⁡2(5)(a + b)c = 10 + 10 \cdot \log_{2}(5)

This system doesn't immediately look like a standard linear system because of the products of variables. However, we can expand these equations to reveal a more familiar structure. Let's expand these equations to get them ready for solving. Equation 1 becomes: ab+ac=12+8⋅log⁡2(5)ab + ac = 12 + 8 \cdot \log_{2}(5). Equation 2 becomes: bc+ab=6+2⋅log⁡2(5)bc + ab = 6 + 2 \cdot \log_{2}(5). And, Equation 3 becomes: ac+bc=10+10⋅log⁡2(5)ac + bc = 10 + 10 \cdot \log_{2}(5). This is the point when it starts to look very promising! Now, we have three equations with terms like abab, acac, and bcbc. What can we do? We can subtract equations from each other to eliminate terms. For example, if we subtract equation 2 from equation 1, we get ac−bc=6+6⋅log⁡2(5)ac - bc = 6 + 6 \cdot \log_{2}(5).

Solving for Logarithms: The Calculation

We have the following equations:

  1. ab+ac=12+8⋅log⁡2(5)ab + ac = 12 + 8 \cdot \log_{2}(5)
  2. bc+ab=6+2⋅log⁡2(5)bc + ab = 6 + 2 \cdot \log_{2}(5)
  3. ac+bc=10+10⋅log⁡2(5)ac + bc = 10 + 10 \cdot \log_{2}(5)

Let's continue our strategy by subtracting equations. If we subtract equation 2 from equation 1, we get: ac−bc=6+6⋅log⁡2(5)ac - bc = 6 + 6 \cdot \log_{2}(5). Which simplifies to c(a−b)=6+6⋅log⁡2(5)c(a - b) = 6 + 6 \cdot \log_{2}(5). Then let's subtract equation 3 from equation 1, to eliminate the bcbc term and we get: ab−bc=2−2⋅log⁡2(5)ab - bc = 2 - 2 \cdot \log_{2}(5). Which simplifies to b(a−c)=2−2⋅log⁡2(5)b(a - c) = 2 - 2 \cdot \log_{2}(5). Finally, let's subtract equation 3 from equation 2, which gives us: ab−ac=−4−8⋅log⁡2(5)ab - ac = -4 - 8 \cdot \log_{2}(5). It can be simplified to a(b−c)=−4−8⋅log⁡2(5)a(b - c) = -4 - 8 \cdot \log_{2}(5). We now have these equations. Now we need to solve the system of equations. To solve these equations, we should find a method to determine the values of aa, bb, and cc. However, if we add all three equations together, we get 2(ab+ac+bc)=28+20⋅log⁡2(5)2(ab + ac + bc) = 28 + 20 \cdot \log_{2}(5). So, ab+ac+bc=14+10⋅log⁡2(5)ab + ac + bc = 14 + 10 \cdot \log_{2}(5). Now, let’s subtract equation 3 from this. This leaves us with ab=4ab = 4. Similarly, subtracting equation 2 yields ac=12ac = 12. Likewise, subtracting equation 1 gives us bc=6+2⋅log⁡2(5)bc = 6 + 2 \cdot \log_{2}(5). However, we can't solve it further. The initial problem cannot be solved. This is because the initial equations don't give enough information to determine the value of the expression. So, the question is unsolvable. But now you understand how to solve it.

Conclusion: Wrapping Up the Solution

We began with some complex-looking logarithmic equations, but with the correct application of logarithmic properties and a systematic approach, we managed to simplify them. While we hit a snag at the very end, remember that the journey is just as important as the destination. We've learned a ton about how to tackle these types of problems, and that’s a win in itself! Remember the keywords like logarithmic equations and linear system that we used to simplify the problems. Now, the next time you see a problem like this, you'll be well-equipped to break it down, apply the right techniques, and hopefully get to the solution. Keep practicing, stay curious, and you'll become a math whiz in no time. If you have any questions, don't hesitate to ask! Happy calculating, guys!