Solving Simultaneous Logarithmic Equations Find X And Y

Hey guys! Today, we're diving deep into the fascinating world of simultaneous logarithmic equations. If you've ever felt a little intimidated by these problems, don't worry, you're not alone. We're going to break down the steps involved in solving them, making it super clear and easy to understand. So, buckle up, and let's get started!

Understanding Logarithmic Equations

Before we jump into solving simultaneous equations, let's quickly recap what logarithmic equations are all about. Remember, a logarithm is just another way of expressing an exponent. The equation $\log_b a = c$ is equivalent to $b^c = a$. Here, $b$ is the base of the logarithm, $a$ is the argument, and $c$ is the exponent. Think of it like this: the logarithm tells you what power you need to raise the base to, in order to get the argument.

When dealing with logarithmic equations, it's crucial to understand the properties of logarithms. These properties are your best friends when it comes to simplifying and solving these equations. Some key properties include:

1. Product Rule: $\log_b (mn) = \log_b m + \log_b n$
2. Quotient Rule: $\log_b (m/n) = \log_b m - \log_b n$
3. Power Rule: $\log_b (m^p) = p \log_b m$
4. Change of Base Formula: $\log_b a = \frac{\log_c a}{\log_c b}$

These properties allow us to manipulate logarithmic expressions, combine or separate terms, and change the base of the logarithm, which can be incredibly useful when solving equations. In this article, we will solve the equations, giving the values of $x$ and $y$ to 3 significant figures. So, understanding these rules is half the battle, guys!

Setting Up the Problem

Now, let's tackle the problem at hand. We've got a system of two equations:

$3 \log_2 x + 4 \log_3 y = 10$ ... (1) $\log_2 x - 2 \log_3 y = 1$ ... (2)

The goal here is to find the values of $x$ and $y$ that satisfy both equations simultaneously. Sounds like a mission, right? But don't worry, we've got a plan. The key to solving simultaneous equations is to eliminate one of the variables, so we're left with a single equation in a single variable. We can then solve for that variable and substitute the result back into one of the original equations to find the other variable.

Looking at our equations, we can see that the coefficients of $\log_3 y$ are 4 and -2. This gives us a great opportunity to eliminate $\log_3 y$. If we multiply equation (2) by 2, the coefficient of $\log_3 y$ will become -4, which is the negative of the coefficient in equation (1). When we add the two equations together, the $\log_3 y$ terms will cancel out, leaving us with an equation in terms of $\log_2 x$ only. This is a classic technique in solving simultaneous equations, and it's super effective. This sets us up perfectly to isolate and solve for our variables.

Elimination Method in Action

Okay, let's get our hands dirty and apply the elimination method. As we discussed, we're going to multiply equation (2) by 2:

$2(\log_2 x - 2 \log_3 y) = 2(1)$ $2 \log_2 x - 4 \log_3 y = 2$ ... (3)

Now we have equation (3), and we can add it to equation (1). This is where the magic happens! When we add the equations, the terms involving $\log_3 y$ will disappear, making our lives much easier. So, let's add equation (1) and equation (3):

$(3 \log_2 x + 4 \log_3 y) + (2 \log_2 x - 4 \log_3 y) = 10 + 2$ $3 \log_2 x + 2 \log_2 x = 12$ $5 \log_2 x = 12$

See how the $\log_3 y$ terms neatly canceled out? We're now left with a single equation involving only $\log_2 x$. This is a major step forward! Now, we can easily solve for $\log_2 x$ by dividing both sides of the equation by 5:

$\log_2 x = \frac{12}{5}$

We've successfully isolated $\log_2 x$. The next step is to find the value of $x$ itself. We're getting closer to the solution, guys! This is what we are aiming for when dealing with logarithmic expressions.

Solving for x

Now that we have $\log_2 x = \frac{12}{5}$, we need to find the value of $x$. Remember, the logarithm is just an exponent in disguise. To get $x$ by itself, we need to rewrite the logarithmic equation in exponential form. The equation $\log_2 x = \frac{12}{5}$ is equivalent to:

$x = 2^{\frac{12}{5}}$

This is because the logarithm base 2 of $x$ equals $\frac{12}{5}$, which means 2 raised to the power of $\frac{12}{5}$ gives us $x$. To find the numerical value of $x$, we can use a calculator. Make sure you know how to use the exponent function on your calculator – it's a crucial skill for solving these types of problems. When you plug $2^{\frac{12}{5}}$ into your calculator, you should get approximately:

$x \approx 5.278031643$

But, the question asks for the value of $x$ to 3 significant figures. So, we need to round our result accordingly. The first three significant figures are 5, 2, and 7. The next digit is 8, which is greater than or equal to 5, so we round up the last significant figure. Therefore:

$x \approx 5.28$

We've found the value of $x$ to 3 significant figures! That's a big accomplishment. But remember, we're not done yet. We still need to find the value of $y$. To do this, we'll substitute our value of $x$ back into one of the original equations and solve for $y$. Keep going, guys! We're on a roll.

Solving for y

Alright, we've nailed down the value of $x$, which is approximately 5.28. Now, it's time to hunt down the value of $y$. To do this, we'll substitute our calculated value of $x$ back into one of the original equations. It doesn't matter which equation we choose, but it's often easier to pick the simpler one. In this case, equation (2) looks a bit less intimidating:

$\log_2 x - 2 \log_3 y = 1$

Substitute $x \approx 5.28$ into this equation:

$\log_2 (5.28) - 2 \log_3 y = 1$

Now, we need to isolate the term involving $y$. First, let's calculate $\log_2 (5.28)$. You'll need a calculator for this, and you might need to use the change of base formula if your calculator doesn't have a $\log_2$ function. The change of base formula is:

$\log_b a = \frac{\log_c a}{\log_c b}$

So, we can rewrite $\log_2 (5.28)$ as $\frac{\log_{10} (5.28)}{\log_{10} (2)}$ or $\frac{\ln (5.28)}{\ln (2)}$. Using a calculator, we find:

$\log_2 (5.28) \approx 2.397$

Now, substitute this value back into our equation:

$2.397 - 2 \log_3 y = 1$

Next, let's isolate the term with $\log_3 y$. Subtract 2.397 from both sides:

$-2 \log_3 y = 1 - 2.397$ $-2 \log_3 y = -1.397$

Now, divide both sides by -2:

$\log_3 y = \frac{-1.397}{-2}$ $\log_3 y \approx 0.6985$

We're almost there! We've isolated $\log_3 y$, and now we just need to find $y$. Just like we did with $x$, we'll rewrite this logarithmic equation in exponential form:

$y = 3^{0.6985}$

Use your calculator to find the value of $3^{0.6985}$:

$y \approx 2.143$

Finally, we need to round this to 3 significant figures. The first three significant figures are 2, 1, and 4. The next digit is 3, which is less than 5, so we don't need to round up. Therefore:

$y \approx 2.14$

We've found the value of $y$ to 3 significant figures! High five! We've successfully navigated the logarithmic maze and emerged victorious.

Final Answer

We've done it, guys! We've solved the simultaneous logarithmic equations and found the values of $x$ and $y$ to 3 significant figures. Let's summarize our results:

$x \approx 5.28$ $y \approx 2.14$

These are the solutions to the system of equations. We tackled a potentially tricky problem by breaking it down into smaller, manageable steps. We used the properties of logarithms, the elimination method, and a bit of calculator magic to arrive at our answer. Remember, the key to solving these types of problems is practice. The more you work with logarithms, the more comfortable you'll become with them. So, keep practicing, and you'll become a logarithm master in no time! You can try out more exercises on logarithmic expressions to improve your skills.

Conclusion

Solving simultaneous logarithmic equations might seem daunting at first, but with a clear understanding of the properties of logarithms and a systematic approach, you can conquer any problem. Remember to break down the problem into steps, use the elimination or substitution method to reduce the number of variables, and don't forget to use your calculator wisely. And most importantly, practice makes perfect!

So, there you have it! A comprehensive guide to solving simultaneous logarithmic equations. I hope this has been helpful and that you're now feeling more confident in your ability to tackle these types of problems. Keep up the great work, and I'll see you next time!